C（VS2005）

Question

挑战

这是挑战（我自己的发明，但如果以前出现在网络的其他地方，我也不会感到惊讶。）

编写一个单一的函数   论证是一个   字符串表示一个简单的   数学表达和评估   它作为浮点值。一个   “简单表达”可以包括任何一个   以下：积极或消极   十进制数字， + ， - ， * ， / ，（，）即可。   表达式使用（正常）infix notation。   应该对运营商进行评估   他们出现的顺序，即不   BODMAS，   虽然括号应该是正确的   当然，观察到了。该函数应该返回   任何可能表达式的正确结果   这种形式。但是，该功能没有   处理格式错误的表达式（即语法错误的表达式）。

表达式示例：
1 + 3 / -8                            = -0.5       (No BODMAS)
2*3*4*5+99                            = 219
4 * (9 - 4) / (2 * 6 - 2) + 8         = 10
1 + ((123 * 3 - 69) / 100)            = 4
2.45/8.5*9.27+(5*0.0023)              = 2.68...

规则

我预计会有某种形式的“作弊”/狡猾，所以请让我预先反对它！通过作弊，我指的是在动态语言（如JavaScript或PHP）中使用eval或等效函数，或者同时编译和执行代码。（我认为我的“无BODMAS”的规格已经基本保证了这一点。）除此之外，没有任何限制。我在这里预计会有一些Regex解决方案，但是看到的不仅仅是那个很好。

现在，我主要对C＃/ .NET解决方案感兴趣，但是任何其他语言都是完全可以接受的（特别是F＃和Python用于功能/混合方法）。我还没有决定是否要接受最短或最巧妙的解决方案（至少对于语言而言）作为答案，但我欢迎任何语言的任何形式的解决方案，除了我上面禁止的内容！

我的解决方案

我现在发布了我的C＃解决方案here（403个字符）。 更新： My new solution在 294个字符中显着击败了旧版本，借助一些可爱的正则表达式！我怀疑这会被一些语言轻松打败，语法较轻（尤其是功能/动态语言），并且已被证明是正确的，但我很好奇是否有人能在C＃中击败它。

更新

我已经看到了一些非常狡猾的解决方案。感谢所有发布过的人。虽然我还没有测试过任何一个，但我会相信人们，并假设他们至少可以使用所有给定的例子。

仅仅是为了注释，重新入门（即线程安全）不是对该功能的要求，尽管这是一个奖励。

格式

请以下列格式发布所有答案，以便于比较：

语言

字符数：???

完全混淆的功能：
(code here)
清除/半混淆功能：
(code here)
关于算法/聪明的快捷方式的任何注释。

Answer 1

汇编

427字节

混淆，用优秀的A86汇总成.com可执行文件：

dd 0db9b1f89h, 081bee3h, 0e8af789h, 0d9080080h, 0bdac7674h, 013b40286h
dd 07400463ah, 0ccfe4508h, 08ce9f675h, 02fc8000h, 013b0057eh, 0feaac42ah
dd 0bedf75c9h, 0ba680081h, 04de801h, 04874f73bh, 04474103ch, 0e8e8b60fh
dd 08e8a003fh, 0e880290h, 0de0153h, 08b57e6ebh, 0d902a93eh, 046d891dh
dd 08906c783h, 05f02a93eh, 03cffcee8h, 057197510h, 02a93e8bh, 08b06ef83h
dd 05d9046dh, 02a93e89h, 03bc9d95fh, 0ac0174f7h, 074f73bc3h, 0f3cac24h
dd 0eed9c474h, 0197f0b3ch, 07cc4940fh, 074f73b09h, 0103cac09h, 0a3ce274h
dd 0e40a537eh, 0e0d90274h, 02a3bac3h, 021cd09b4h, 03e8b20cdh, 0ff8102a9h
dd 0ed7502abh, 0474103ch, 0e57d0b3ch, 0be02a3bfh, 014d903a3h, 0800344f6h
dd 02db00574h, 0d9e0d9aah, 0d9029f2eh, 0bb34dfc0h, 08a0009h, 01c75f0a8h
dd 020750fa8h, 0b0f3794bh, 021e9aa30h, 0de607400h, 08802990eh, 0de07df07h
dd 0c392ebc1h, 0e8c0008ah, 0aa300404h, 0f24008ah, 04baa3004h, 02eb0ee79h
dd 03005c6aah, 0c0d90ab1h, 0e9defcd9h, 02a116deh, 0e480e0dfh, 040fc8045h
dd 0ede1274h, 0c0d90299h, 015dffcd9h, 047300580h, 0de75c9feh, 0303d804fh
dd 03d80fa74h, 04f01752eh, 0240145c6h, 0dfff52e9h, 0d9029906h, 0f73b025fh
dd 03caca174h, 07fed740ah, 0df07889ah, 0277d807h, 047d9c1deh, 0990ede02h
dd 025fd902h, 03130e0ebh, 035343332h, 039383736h, 02f2b2d2eh, 02029282ah
dd 0e9000a09h, 07fc9f9c1h, 04500000fh, 0726f7272h
db 024h, 0abh, 02h

编辑：未经过模糊处理的来源：

        mov [bx],bx
        finit
        mov si,81h
        mov di,si
        mov cl,[80h]
        or cl,bl
        jz ret
    l1:
        lodsb
        mov bp,d1
        mov ah,19
    l2:
        cmp al,[bp]
        je l3
        inc bp
        dec ah
        jne l2
        jmp exit
    l3:
        cmp ah,2
        jle l4
        mov al,19
        sub al,ah
        stosb
    l4:
        dec cl
        jnz l1
        mov si,81h
        push done

    decode:
    l5:
        call l7
    l50:
        cmp si,di
        je ret
        cmp al,16
        je ret
        db 0fh, 0b6h, 0e8h ; movzx bp,al
        call l7
        mov cl,[bp+op-11]
        mov byte ptr [sm1],cl
        db 0deh
    sm1:db ?
        jmp l50

    open:
        push di
        mov di,word ptr [s]
        fstp dword ptr [di]
        mov [di+4],bp
        add di,6
        mov word ptr [s],di
        pop di
        call decode
        cmp al,16
        jne ret
        push di
        mov di,word ptr [s]
        sub di,6
        mov bp,[di+4]
        fld dword ptr [di]
        mov word ptr [s],di
        pop di
        fxch st(1)
        cmp si,di
        je ret
        lodsb
        ret



    l7: cmp si,di
        je exit
        lodsb
        cmp al,15
        je open
        fldz
        cmp al,11
        jg exit
        db 0fh, 94h, 0c4h ; sete ah 
        jl l10
    l9:
        cmp si,di
        je l12
        lodsb
        cmp al,16
        je ret
    l10:
        cmp al,10
        jle l12i

    l12:
        or ah,ah
        je l13
        fchs
    l13:
        ret

    exit:
        mov dx,offset res
        mov ah,9
        int 21h
        int 20h

    done:
        mov di,word ptr [s]
        cmp di,(offset s)+2
        jne exit
        cmp al,16
        je ok
        cmp al,11
        jge exit
    ok:
        mov di,res
        mov si,res+100h
        fst dword ptr [si]
        test byte ptr [si+3],80h
        jz pos
        mov al,'-'
        stosb
        fchs
    pos:
        fldcw word ptr [cw]
        fld st(0)
        fbstp [si]
        mov bx,9
    l1000:
        mov al,[si+bx]
        test al,0f0h
        jne startu
        test al,0fh
        jne startl
        dec bx
        jns l1000
        mov al,'0'
        stosb
        jmp frac

    l12i:
        je l11
        fimul word ptr [d3]
        mov [bx],al
        fild word ptr [bx]
        faddp
        jmp l9
        ret

    startu:
        mov al,[si+bx]
        shr al,4
        add al,'0'
        stosb
    startl:
        mov al,[si+bx]
        and al,0fh
        add al,'0'
        stosb
        dec bx
        jns startu

    frac:
        mov al,'.'
        stosb
        mov byte ptr [di],'0'
        mov cl,10
        fld st(0)
        frndint
    frac1:  
        fsubp st(1)
        ficom word ptr [zero]
        fstsw ax
        and ah,045h
        cmp ah,040h
        je finished
        fimul word ptr [d3]
        fld st(0)
        frndint
        fist word ptr [di]
        add byte ptr [di],'0'
        inc di
        dec cl
        jnz frac1

    finished:   
        dec di
        cmp byte ptr [di],'0'
        je finished
        cmp byte ptr [di],'.'
        jne f2
        dec di
    f2:
        mov byte ptr [di+1],'$'
    exit2:
        jmp exit


    l11:
        fild word ptr [d3]
        fstp dword ptr [bx+2]
    l111:
        cmp si,di
        je ret
        lodsb
        cmp al,10
        je exit2
        jg ret
        mov [bx],al
        fild word ptr [bx]
        fdiv dword ptr [bx+2]
        faddp
        fld dword ptr [bx+2]
        fimul word ptr [d3]
        fstp dword ptr [bx+2]
        jmp l111


    d1: db '0123456789.-+/*()', 32, 9
    d3: dw 10
    op: db 0e9h, 0c1h, 0f9h, 0c9h
    cw: dw 0f7fh
    zero: dw 0
    res:db 'Error$'
    s:  dw (offset s)+2

Answer 2

Perl（无评估）

字符数： ~~167~~ 106 （106字符版本见下文）

完全混淆的功能:(如果将这三行连接成一行，则为167个字符）

sub e{my$_="($_[0])";s/\s//g;$n=q"(-?\d++(\.\d+)?+)";
@a=(sub{$1},1,sub{$3*$6},sub{$3+$6},4,sub{$3-$6},6,sub{$3/$6});
while(s:\($n\)|(?<=\()$n(.)$n:$a[7&ord$5]():e){}$_}

清除/反混淆版本：

sub e {
  my $_ = "($_[0])";
  s/\s//g;
  $n=q"(-?\d++(\.\d+)?+)"; # a regex for "number", including capturing groups
                           # q"foo" in perl means the same as 'foo'
                           # Note the use of ++ and ?+ to tell perl
                           # "no backtracking"

  @a=(sub{$1},             # 0 - no operator found
      1,                   # placeholder
      sub{$3*$6},          # 2 - ord('*') = 052
      sub{$3+$6},          # 3 - ord('+') = 053
      4,                   # placeholder
      sub{$3-$6},          # 5 - ord('-') = 055
      6,                   # placeholder
      sub{$3/$6});         # 7 - ord('/') = 057

  # The (?<=... bit means "find a NUM WHATEVER NUM sequence that happens
  # immediately after a left paren", without including the left
  # paren.  The while loop repeatedly replaces "(" NUM WHATEVER NUM with
  # "(" RESULT and "(" NUM ")" with NUM.  The while loop keeps going
  # so long as those replacements can be made.

  while(s:\($n\)|(?<=\()$n(.)$n:$a[7&ord$5]():e){}

  # A perl function returns the value of the last statement
  $_
}

我最初误读了规则，所以我提交了一个带有“eval”的版本。这是没有它的版本。

当我意识到+，-，/和*的字符代码中的最后一个八进制数字不同时，最新的洞察力出现了， ord(undef)为0.这使我可以将调度表@a设置为数组，并只调用7 & ord($3)位置的代码。

有一个显而易见的地方可以削减一个角色 - 将q""更改为'' - 但这会使切割和粘贴到外壳中变得更加困难。

更短

字符数： ~~124~~ 106

考虑到ephemient的编辑，现在已经减少到124个字符:(将两行合并为一行）

sub e{$_=$_[0];s/\s//g;$n=q"(-?\d++(\.\d+)?+)";
1while s:\($n\)|$n(.)$n:($1,1,$3*$6,$3+$6,4,$3-$6,6,$6&&$3/$6)[7&ord$5]:e;$_}

更短

字符数： ~~110~~ 106

下面的红宝石解决方案让我更进一步，但我无法达到它的104个字符：

sub e{($_)=@_;$n='( *-?[.\d]++ *)';
s:\($n\)|$n(.)$n:(($1,$2-$4,$4&&$2/$4,$2*$4,$2+$4)x9)[.8*ord$3]:e?e($_):$_}

我不得不放弃并使用''。这个ruby send技巧对这个问题非常有用。

从石头中挤水

字符数：106

避免零除检查的小挫折。

sub e{($_)=@_;$n='( *-?[.\d]++ *)';
s:\($n\)|$n(.)$n:($1,0,$2*$4,$2+$4,0,$2-$4)[7&ord$3]//$2/$4:e?e($_):$_}

以下是此功能的测试工具：

perl -le 'sub e{($_)=@_;$n='\''( *-?[.\d]++ *)'\'';s:\($n\)|$n(.)$n:($1,0,$2*$4,$2+$4,0,$2-$4)[7&ord$3]//$2/$4:e?e($_):$_}' -e 'print e($_) for @ARGV' '1 + 3' '1 + ((123 * 3 - 69) / 100)' '4 * (9 - 4) / (2 * 6 - 2) + 8' '2*3*4*5+99' '2.45/8.5*9.27+(5*0.0023) ' '1 + 3 / -8'

Answer 3

C（VS2005）

字符数：1360

滥用预处理器和警告以获得有趣的代码布局（向下滚动以查看）：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define b main
#define c(a) b(a,0)
#define d -1
#define e -2
#define g break
#define h case
#define hh h
#define hhh h
#define w(i) case i
#define i return
#define j switch
#define k float
#define l realloc
#define m sscanf
#define n int _
#define o char
#define t(u) #u
#define q(r) "%f" t(r)  "n"
#define s while
#define v default
#define ex exit
#define W printf
#define x fn()
#define y strcat
#define z strcpy
#define Z strlen

char*p    =0    ;k    *b    (n,o**    a){k*f
;j(_){    hh   e:     i*    p==40?    (++p,c
(d        ))  :(      f=        l(        0,
4)        ,m (p       ,q        (%        ),
f,&_),    p+=_        ,f       );        hh
d:f=c(    e);s        (1      ){        j(
    *p    ++ ){       hh     0:        hh
    41    :i  f;      hh    43        :*
f+=*c(    e)   ;g     ;h    45:*f=    *f-*c(
e);g;h    42    :*    f=    *f**c(    e);g;h

47:*f      /=*c      (e);     g;   v:    c(0);}
}w(1):    if(p&&    printf    (q  ((     "\\"))
,*  c(    d)  ))    g;  hh    0: ex      (W
(x  ))    ;v  :p    =(        p?y:       z)(l(p
,Z(1[     a]  )+    (p        ?Z(p           )+
1:1))     ,1  [a    ])  ;b    (_ -1          ,a
+1  );    g;  }i    0;};fn    ()  {n     =42,p=
43  ;i     "Er"      "ro"     t(   r)    "\n";}

Answer 4

红宝石

字符数：103

N='( *-?[\d.]+ *)'
def e x
x.sub!(/\(#{N}\)|#{N}([^.\d])#{N}/){$1or(e$2).send$3,e($4)}?e(x):x.to_f
end

这是The Wicked Flea解决方案的~~非递归~~版本。括号内的子表达式是自下而上而不是自上而下的评估。

编辑：将'while'转换为条件+尾递归已保存了几个字符，因此它不再是非递归的（尽管递归在语义上是不必要的。）

编辑：借用丹尼尔·马丁关于合并正则表达式的想法可以节省另外11个字符！

编辑：这种递归比我原先想象的更有用！如果x.to_f恰好包含一个数字，e(x)可以重写为x。

修改：使用“or”代替“||”可以删除一对括号。

长版：

# Decimal number, as a capturing group, for substitution
# in the main regexp below.
N='( *-?[\d.]+ *)'

# The evaluation function
def e(x)
  matched = x.sub!(/\(#{N}\)|#{N}([^\d.])#{N}/) do
    # Group 1 is a numeric literal in parentheses.  If this is present then
    # just return it.
    if $1
      $1
    # Otherwise, $3 is an operator symbol and $2 and $4 are the operands
    else
      # Recursively call e to parse the operands (we already know from the
      # regexp that they are numeric literals, and this is slightly shorter
      # than using :to_f)
      e($2).send($3, e($4))
      # We could have converted $3 to a symbol ($3.to_s) or converted the
      # result back to string form, but both are done automatically anyway
    end
  end
  if matched then
    # We did one reduction. Now recurse back and look for more.
    e(x)
  else
    # If the string doesn't look like a non-trivial expression, assume it is a
    # string representation of a real number and attempt to parse it
    x.to_f
  end
end

Answer 5

的Haskell

字符数：182

没有尝试聪明，只是一些压缩：4行，312字节。

import Data.Char;import Text.ParserCombinators.Parsec
q=either(error.show)id.runParser t id"".filter(' '/=);t=do
s<-getState;a<-fmap read(many1$oneOf".-"<|>digit)<|>between(char '('>>setState id)(char ')'>>setState s)t
option(s a)$choice(zipWith(\c o->char c>>return(o$s a))"+-*/"[(+),(-),(*),(/)])>>=setState>>t

现在，真正进入高尔夫精神，3线和182字节：

q=snd.(`e`id).filter(' '/=)
e s c|[(f,h)]<-readsPrec 0 s=g h(c f);e('(':s)c=g h(c f)where(')':h,f)=e s id
g('+':h)=e h.(+);g('-':h)=e h.(-);g('*':h)=e h.(*);g('/':h)=e h.(/);g h=(,)h

分解：

-- Strip spaces from the input, evaluate with empty accumulator,
-- and output the second field of the result.
q :: String -> Double
q = snd . flip eval id . filter (not . isSpace)

-- eval takes a string and an accumulator, and returns
-- the final value and what’s left unused from the string.
eval :: (Fractional a, Read a) => String -> (a -> a) -> (String, a)

-- If the beginning of the string parses as a number, add it to the accumulator,
-- then try to read an operator and further.
eval str accum | [(num, rest)] <- readsPrec 0 str = oper rest (accum num)

-- If the string starts parentheses, evaluate the inside with a fresh
-- accumulator, and continue after the closing paren.
eval ('(':str) accum = oper rest (accum num) where (')':rest, num) = eval str id

-- oper takes a string and current value, and tries to read an operator
-- to apply to the value.  If there is none, it’s okay.
oper :: (Fractional a, Read a) => String -> a -> (String, a)

-- Handle operations by giving eval a pre-seeded accumulator.
oper ('+':str) num = eval str (num +)
oper ('-':str) num = eval str (num -)
oper ('*':str) num = eval str (num *)
oper ('/':str) num = eval str (num /)

-- If there’s no operation parsable, just return.
oper str num = (str, num)

Answer 6

Visual Basic.NET

字符数：9759

我自己更喜欢投球手。

注意：不考虑嵌套括号。此外，未经测试，但我很确定它有效。

Imports Microsoft.VisualBasic
Imports System.Text
Imports System.Collections.Generic
Public Class Main
Public Shared Function DoArithmaticFunctionFromStringInput(ByVal MathematicalString As String) As Double
    Dim numberList As New List(Of Number)
    Dim operationsList As New List(Of IOperatable)
    Dim currentNumber As New Number
    Dim currentParentheticalStatement As New Parenthetical
    Dim isInParentheticalMode As Boolean = False
    Dim allCharactersInString() As Char = MathematicalString.ToCharArray
    For Each mathChar In allCharactersInString
        If mathChar = Number.ZERO_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Number.ONE_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Number.TWO_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Number.THREE_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Number.FOUR_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Number.FIVE_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Number.SIX_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Number.SEVEN_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Number.EIGHT_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Number.NINE_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Number.DECIMAL_POINT_STRING_REPRESENTATION Then
            currentNumber.UpdateNumber(mathChar)
        ElseIf mathChar = Addition.ADDITION_STRING_REPRESENTATION Then
            Dim addition As New Addition

            If Not isInParentheticalMode Then
                operationsList.Add(addition)
                numberList.Add(currentNumber)
            Else
                currentParentheticalStatement.AllNumbers.Add(currentNumber)
                currentParentheticalStatement.AllOperators.Add(addition)
            End If

            currentNumber = New Number
        ElseIf mathChar = Number.NEGATIVE_NUMBER_STRING_REPRESENTATION Then
            If currentNumber.StringOfNumbers.Length > 0 Then
                currentNumber.UpdateNumber(mathChar)

                Dim subtraction As New Addition
                If Not isInParentheticalMode Then
                    operationsList.Add(subtraction)
                    numberList.Add(currentNumber)
                Else
                    currentParentheticalStatement.AllNumbers.Add(currentNumber)
                    currentParentheticalStatement.AllOperators.Add(subtraction)
                End If

                currentNumber = New Number
            Else
                currentNumber.UpdateNumber(mathChar)
            End If
        ElseIf mathChar = Multiplication.MULTIPLICATION_STRING_REPRESENTATION Then
            Dim multiplication As New Multiplication

            If Not isInParentheticalMode Then
                operationsList.Add(multiplication)
                numberList.Add(currentNumber)
            Else
                currentParentheticalStatement.AllNumbers.Add(currentNumber)
                currentParentheticalStatement.AllOperators.Add(multiplication)
            End If
            currentNumber = New Number
        ElseIf mathChar = Division.DIVISION_STRING_REPRESENTATION Then
            Dim division As New Division

            If Not isInParentheticalMode Then
                operationsList.Add(division)
                numberList.Add(currentNumber)
            Else
                currentParentheticalStatement.AllNumbers.Add(currentNumber)
                currentParentheticalStatement.AllOperators.Add(division)
            End If
            currentNumber = New Number
        ElseIf mathChar = Parenthetical.LEFT_PARENTHESIS_STRING_REPRESENTATION Then
            isInParentheticalMode = True
        ElseIf mathChar = Parenthetical.RIGHT_PARENTHESIS_STRING_REPRESENTATION Then
            currentNumber = currentParentheticalStatement.EvaluateParentheticalStatement
            numberList.Add(currentNumber)
            isInParentheticalMode = False
        End If
    Next

    Dim result As Double = 0
    Dim operationIndex As Integer = 0
    For Each numberOnWhichToPerformOperations As Number In numberList
        result = operationsList(operationIndex).PerformOperation(result, numberOnWhichToPerformOperations)
        operationIndex = operationIndex + 1
    Next

    Return result

End Function
Public Class Number
    Public Const DECIMAL_POINT_STRING_REPRESENTATION As Char = "."
    Public Const NEGATIVE_NUMBER_STRING_REPRESENTATION As Char = "-"
    Public Const ZERO_STRING_REPRESENTATION As Char = "0"
    Public Const ONE_STRING_REPRESENTATION As Char = "1"
    Public Const TWO_STRING_REPRESENTATION As Char = "2"
    Public Const THREE_STRING_REPRESENTATION As Char = "3"
    Public Const FOUR_STRING_REPRESENTATION As Char = "4"
    Public Const FIVE_STRING_REPRESENTATION As Char = "5"
    Public Const SIX_STRING_REPRESENTATION As Char = "6"
    Public Const SEVEN_STRING_REPRESENTATION As Char = "7"
    Public Const EIGHT_STRING_REPRESENTATION As Char = "8"
    Public Const NINE_STRING_REPRESENTATION As Char = "9"

    Private _isNegative As Boolean
    Public ReadOnly Property IsNegative() As Boolean
        Get
            Return _isNegative
        End Get
    End Property
    Public ReadOnly Property ActualNumber() As Double
        Get
            Dim result As String = ""
            If HasDecimal Then
                If DecimalIndex = StringOfNumbers.Length - 1 Then
                    result = StringOfNumbers.ToString
                Else
                    result = StringOfNumbers.Insert(DecimalIndex, DECIMAL_POINT_STRING_REPRESENTATION).ToString
                End If
            Else
                result = StringOfNumbers.ToString
            End If
            If IsNegative Then
                result = NEGATIVE_NUMBER_STRING_REPRESENTATION & result
            End If
            Return CType(result, Double)
        End Get
    End Property
    Private _hasDecimal As Boolean
    Public ReadOnly Property HasDecimal() As Boolean
        Get
            Return _hasDecimal
        End Get
    End Property
    Private _decimalIndex As Integer
    Public ReadOnly Property DecimalIndex() As Integer
        Get
            Return _decimalIndex
        End Get
    End Property
    Private _stringOfNumbers As New StringBuilder
    Public ReadOnly Property StringOfNumbers() As StringBuilder
        Get
            Return _stringOfNumbers
        End Get
    End Property
    Public Sub UpdateNumber(ByVal theDigitToAppend As Char)
        If IsNumeric(theDigitToAppend) Then
            Me._stringOfNumbers.Append(theDigitToAppend)
        ElseIf theDigitToAppend = DECIMAL_POINT_STRING_REPRESENTATION Then
            Me._hasDecimal = True
            Me._decimalIndex = Me._stringOfNumbers.Length
        ElseIf theDigitToAppend = NEGATIVE_NUMBER_STRING_REPRESENTATION Then
            Me._isNegative = Not Me._isNegative
        End If
    End Sub
    Public Shared Function ConvertDoubleToNumber(ByVal numberThatIsADouble As Double) As Number
        Dim numberResult As New Number
        For Each character As Char In numberThatIsADouble.ToString.ToCharArray
            numberResult.UpdateNumber(character)
        Next
        Return numberResult
    End Function
End Class
Public MustInherit Class Operation
    Protected _firstnumber As New Number
    Protected _secondnumber As New Number
    Public Property FirstNumber() As Number
        Get
            Return _firstnumber
        End Get
        Set(ByVal value As Number)
            _firstnumber = value
        End Set
    End Property
    Public Property SecondNumber() As Number
        Get
            Return _secondnumber
        End Get
        Set(ByVal value As Number)
            _secondnumber = value
        End Set
    End Property
End Class
Public Interface IOperatable
    Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double
End Interface
Public Class Addition
    Inherits Operation
    Implements IOperatable
    Public Const ADDITION_STRING_REPRESENTATION As String = "+"
    Public Sub New()

    End Sub
    Public Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double Implements IOperatable.PerformOperation
        Dim result As Double = 0
        result = number1 + number2.ActualNumber
        Return result
    End Function
End Class
Public Class Multiplication
    Inherits Operation
    Implements IOperatable
    Public Const MULTIPLICATION_STRING_REPRESENTATION As String = "*"
    Public Sub New()

    End Sub
    Public Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double Implements IOperatable.PerformOperation
        Dim result As Double = 0
        result = number1 * number2.ActualNumber
        Return result
    End Function
End Class
Public Class Division
    Inherits Operation
    Implements IOperatable
    Public Const DIVISION_STRING_REPRESENTATION As String = "/"
    Public Const DIVIDE_BY_ZERO_ERROR_MESSAGE As String = "I took a lot of time to write this program. Please don't be a child and try to defile it by dividing by zero. Nobody thinks you are funny."
    Public Sub New()

    End Sub
    Public Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double Implements IOperatable.PerformOperation
        If Not number2.ActualNumber = 0 Then
            Dim result As Double = 0
            result = number1 / number2.ActualNumber
            Return result
        Else
            Dim divideByZeroException As New Exception(DIVIDE_BY_ZERO_ERROR_MESSAGE)
            Throw divideByZeroException
        End If
    End Function
End Class
Public Class Parenthetical
    Public Const LEFT_PARENTHESIS_STRING_REPRESENTATION As String = "("
    Public Const RIGHT_PARENTHESIS_STRING_REPRESENTATION As String = ")"
    Private _allNumbers As New List(Of Number)
    Public Property AllNumbers() As List(Of Number)
        Get
            Return _allNumbers
        End Get
        Set(ByVal value As List(Of Number))
            _allNumbers = value
        End Set
    End Property
    Private _allOperators As New List(Of IOperatable)
    Public Property AllOperators() As List(Of IOperatable)
        Get
            Return _allOperators
        End Get
        Set(ByVal value As List(Of IOperatable))
            _allOperators = value
        End Set
    End Property
    Public Sub New()

    End Sub
    Public Function EvaluateParentheticalStatement() As Number
        Dim result As Double = 0
        Dim operationIndex As Integer = 0
        For Each numberOnWhichToPerformOperations As Number In AllNumbers
            result = AllOperators(operationIndex).PerformOperation(result, numberOnWhichToPerformOperations)
            operationIndex = operationIndex + 1
        Next

        Dim numberToReturn As New Number
        numberToReturn = Number.ConvertDoubleToNumber(result)
        Return numberToReturn
    End Function
End Class
End Class

Answer 7

的Python

字符数：237

完全混淆的功能：

from operator import*
def e(s,l=[]):
 if s:l+=list(s.replace(' ','')+')')
 a=0;o=add;d=dict(zip(')*+-/',(0,mul,o,sub,div)));p=l.pop
 while o:
  c=p(0)
  if c=='(':c=e(0)
  while l[0]not in d:c+=p(0)
  a=o(a,float(c));o=d[p(0)]
 return a

清除/半混淆功能：

import operator

def calc(source, stack=[]):
    if source:
        stack += list(source.replace(' ', '') + ')')

    answer = 0

    ops = {
        ')': 0,
        '*': operator.mul,
        '+': operator.add,
        '-': operator.sub,
        '/': operator.div,
    }

    op = operator.add
    while op:
        cur = stack.pop(0)

        if cur == '(':
            cur = calc(0)

        while stack[0] not in ops:
            cur += stack.pop(0)

        answer = op(answer, float(cur))
        op = ops[stack.pop(0)]

    return answer

Answer 8

Fortran 77（gfortran方言，现在支持g77）

字符数： 2059

混淆版本：

      function e(c)
      character*99 c
      character b
      real f(24)                
      integer i(24)             
      nf=0                      
      ni=0                      
 20   nf=kf(0.0,nf,f)
      ni=ki(43,ni,i)         
 30   if (isp(c).eq.1) goto 20
      h=fr(c)
 31   g=fp(nf,f)
      j=ip(ni,i)
      select case(j)
      case (40) 
         goto 20
      case (42)                 
         d=g*h
      case (43)                 
         d=g+h
      case (45)                 
         d=g-h
      case (47)                 
         d=g/h
      end select
 50   nf=kf(d,nf,f)
 60   j=nop(c)
      goto (20, 70, 75, 75, 60, 75, 60, 75) (j-39)
 65   e=fp(nf,f)
      return
 70   h=fp(nf,f)              
      goto 31
 75   ni=ki(j,ni,i)
      goto 30
      end
      function kf(v,n,f)
      real f(24)
      kf=n+1
      f(n+1)=v
      return
      end
      function ki(j,n,i)
      integer i(24)
      ki=n+1
      i(n+1)=j
      return
      end
      function fp(n,f)
      real f(24)
      fp=f(n)
      n=n-1
      return
      end
      function ip(n,i)
      integer i(24)
      ip=i(n)
      n=n-1
      return
      end
      function nop(s)
      character*99 s
      l=1
      do while(s(l:l).eq." ".and.l.lt.99)
         l=l+1
      enddo
      nop=ichar(s(l:l))
      s(l:l)=" "
      return
      end
      function isp(s)
      character*99 s
      isp=0
      l=1
      do while(s(l:l).eq." ".and.l.lt.99)
         l=l+1
      enddo
      isp=41-ichar(s(l:l))
      if (isp.eq.1) s(l:l)=" "
      return
      end
      function fr(s)
      character*99 s
      m=1                      
      n=1                      
      i=1
      do while(i.le.99)
         j=ichar(s(i:i))
         if (j.eq.32) goto 90   
         if (j.ge.48.and.j.lt.58) goto 89
         if (j.eq.43.or.j.eq.45) goto (89,80) m
         if (j.eq.46) goto (83,80) n
 80      exit
 83      n=2
 89      m=2
 90      i=i+1
      enddo
      read(s(1:i-1),*) fr
      do 91 j=1,i-1
         s(j:j)=" "
 91   continue
      return 
      end

清除版本（带支架的3340个字符）

      program infixeval
      character*99 c
      do while (.true.)
         do 10 i=1,99
            c(i:i)=" "
 10      continue
         read(*,"(A99)") c
         f=e(c)
         write(*,*)f
      enddo
      end

      function e(c)
      character*99 c
      character b
      real f(24)                ! value stack
      integer i(24)             ! operator stack
      nf=0                      ! number of items on the value stack
      ni=0                      ! number of items on the operator stack
 20   nf=pushf(0.0,nf,f)
      ni=pushi(43,ni,i)         ! ichar(+) = 43
D     write (*,*) "'",c,"'"
 30   if (isp(c).eq.1) goto 20
      h=fr(c)
D     write (*,*) "'",c,"'"
 31   g=fpop(nf,f)
      j=ipop(ni,i)
D     write(*,*) "Opperate ",g," ",char(j)," ",h
      select case(j)
      case (40) 
         goto 20
      case (42)                 ! "*" 
         d=g*h
      case (43)                 ! "+"
         d=g+h
      case (45)                 ! "-"
         d=g-h
      case (47)                 ! "*"
         d=g/h
      end select
 50   nf=pushf(d,nf,f)
 60   j=nop(c)
D     write(*,*) "Got op: ", char(j)
      goto (20, 70, 75, 75, 60, 75, 60, 75) (j-39)
 65   e=fpop(nf,f)
      return
 70   h=fpop(nf,f)              ! Encountered a "("
      goto 31
 75   ni=pushi(j,ni,i)
      goto 30
      end

c     push onto a real stack
c     OB as kf
      function pushf(v,n,f)
      real f(24)
      pushf=n+1
      f(n+1)=v
D     write(*,*) "Push ", v
      return
      end

c     push onto a integer stack
c     OB as ki
      function pushi(j,n,i)
      integer i(24)
      pushi=n+1
      i(n+1)=j
D     write(*,*) "Push ", char(j)
      return
      end

c     pop from real stack
c     OB as fp
      function fpop(n,f)
      real f(24)
      fpop=f(n)
      n=n-1
D      write (*,*) "Pop ", fpop
      return
      end

c     pop from integer stack
c     OB as ip
      function ipop(n,i)
      integer i(24)
      ipop=i(n)
      n=n-1
D      write (*,*) "Pop ", char(ipop)
      return
      end

c     Next OPerator: returns the next nonws character, and removes it
c     from the string
      function nop(s)
      character*99 s
      l=1
      do while(s(l:l).eq." ".and.l.lt.99)
         l=l+1
      enddo
      nop=ichar(s(l:l))
      s(l:l)=" "
      return
      end

c     IS an open Paren: return 1 if the next non-ws character is "("
c     (also overwrite it with a space. Otherwise return not 1
      function isp(s)
      character*99 s
      isp=0
      l=1
      do while(s(l:l).eq." ".and.l.lt.99)
         l=l+1
      enddo
      isp=41-ichar(s(l:l))
      if (isp.eq.1) s(l:l)=" "
      return
      end

c     Float Read: return the next real number in the string and removes the
c     character
      function fr(s)
      character*99 s
      m=1                      ! No sign (Minus or plus) so far
      n=1                      ! No decimal so far
      i=1
      do while(i.le.99)
         j=ichar(s(i:i))
         if (j.eq.32) goto 90   ! skip spaces
         if (j.ge.48.and.j.lt.58) goto 89
         if (j.eq.43.or.j.eq.45) goto (89,80) m
         if (j.eq.46) goto (83,80) n
c     not part of a number
 80      exit
 83      n=2
 89      m=2
 90      i=i+1
      enddo
      read(s(1:i-1),*) fr
      do 91 j=1,i-1
         s(j:j)=" "
 91   continue
      return 
      end

备注这个编辑过的版本比我的第一次尝试更加邪恶。相同的算法，但现在与goto s的可怕纠结内联。我放弃了合作程序，但现在使用了几种计算分支。已删除所有错误检查和报告，但此版本将从输入中的某些类别的意外字符中静默恢复。这个版本也用g77编译。

主要限制仍然是fortran严格的格式化，长且无处不在的关键字和简单的原语。

Answer 9

C99

字符数：239 （但请参见下文 209 ）

压缩函数：

#define S while(*e==32)++e
#define F float
F strtof();char*e;F v();F g(){S;return*e++-40?strtof(e-1,&e):v();}F v(){F b,a=g();for(;;){S;F o=*e++;if(!o|o==41)return a;b=g();a=o==43?a+b:o==45?a-b:o==42?a*b:a/b;}}F f(char*x){e=x;return v();}

解压缩功能：

float strtof();

char* e;
float v();

float g() {
    while (*e == ' ') ++e;
    return *e++ != '(' ? strtof(e-1, &e) : v();
}

float v() {
    float b, a = g();
    for (;;) {
        while (*e == ' ') ++e;
        float op = *e++;
        if (op == 0 || op == ')') return a;
        b = g();
        a = op == '+' ? a + b : op == '-' ? a - b : op == '*' ? a * b : a / b;
    }
}

float eval(char* x) {
    e = x;
    return v();
}

功能不可重复。

来自Chris Lutz的编辑：我讨厌践踏另一个人的代码，但这里有 209 - 字符版本：

#define S for(;*e==32;e++)
#define X (*e++-40?strtof(e-1,&e):v())
float strtof();char*e;float v(){float o,a=X;for(;;){S;o=*e++;if(!o|o==41)return a;S;a=o-43?o-45?o-42?a/X:a*X:a-X:a+X;}}
#define f(x) (e=x,v())

可读（好吧，不太可读，但解压缩）：

float strtof();
char *e;
float v() {
    float o, a = *e++ != '(' ? strtof(e - 1, &e) : v();
    for(;;) {
        for(; *e == ' '; e++);
        o = *e++;
        if(o == 0 || o==')') return a;
        for(; *e == ' '; e++);
        // I have no idea how to properly indent nested conditionals
        // and this is far too long to fit on one line.
        a = o != '+' ?
          o != '-' ?
            o != '*' ?
              a / (*e++ != '(' ? strtof(e - 1, &e) : v()) :
              a * (*e++ != '(' ? strtof(e - 1, &e) : v()) :
            a - (*e++ != '(' ? strtof(e - 1, &e) : v()) :
          a + (*e++ != '(' ? strtof(e - 1, &e) : v());
      }
}
#define f(x) (e = x, v())

是的，f()是一个宏，不是一个函数，但它有效。可读版本有一些逻辑被重写但没有重新排序（如o != '+'而不是o - '+'），但另外只是另一个的缩进（和预处理）版本。我一直在尝试将if(!o|o==41)return a;部分简化为for()循环，但它永远不会缩短。我仍然相信它可以做到，但我已经完成了打高尔夫球。如果我再处理这个问题，它将在the language that must not be named。

Answer 10

Common Lisp

（SBCL）
字符数：251

(defun g(e)(if(numberp e)e(let((m (g (pop e)))(o(loop for x in e by #'cddr collect x))(n(loop for x in (cdr e)by #'cddr collect (g x))))(mapcar(lambda(x y)(setf m(apply x(list m y))))o n)m)))(defun w(e)(g(read-from-string(concatenate'string"("e")"))))

正确版本（387个字符）：

(defun wrapper (exp) (golf-eval (read-from-string (concatenate 'string "(" exp ")"))))

(defun golf-eval (exp)
 (if (numberp exp)
     exp
   (let ((mem (golf-eval (pop exp)))
     (op-list (loop for x in exp by #'cddr collect x))
     (num-list (loop for x in (cdr exp) by #'cddr collect (golf-eval x))))
    (mapcar (lambda (x y) (setf mem (apply x (list mem y)))) op-list num-list)
    mem)))

输入是w()形式，它接受一个字符串参数。它使用nums /操作数和运算符在N O N O N模式中的技巧，并递归地计算所有操作数，因此嵌套非常便宜。 ;）

Answer 11

JavaScript（不兼容IE）

字符数：268/260

完全混淆的功能：

function e(x){x=x.replace(/ /g,'')+')'
function P(n){return x[0]=='('?(x=x.substr(1),E()):(n=/^[-+]?[\d.]+/(x)[0],x=x.substr(n.length),+n)}function E(a,o,b){a=P()
for(;;){o=x[0]
x=x.substr(1)
if(o==')')return a
b=P()
a=o=='+'?a+b:o=='-'?a-b:o=='*'?a*b:a/b}}return E()}

或者，在JavaScript 1.8（Firefox 3+）中，您可以使用表达式闭包来保存几个字符：

e=function(x,P,E)(x=x.replace(/ /g,'')+')',P=function(n)(x[0]=='('?(x=x.substr(1),E()):(n=/^[-+]?[\d.]+/(x)[0],x=x.substr(n.length),+n)),E=function(a,o,b){a=P()
for(;;){o=x[0]
x=x.substr(1)
if(o==')')return a
b=P()
a=o=='+'?a+b:o=='-'?a-b:o=='*'?a*b:a/b}},E())

清除/半混淆功能：

function evaluate(x) {
    x = x.replace(/ /g, "") + ")";
    function primary() {
        if (x[0] == '(') {
            x = x.substr(1);
            return expression();
        }

        var n = /^[-+]?\d*\.?\d*/.exec(x)[0];
        x = x.substr(n.length);
        return +n;
    }

    function expression() {
        var a = primary();
        for (;;) {
            var operator = x[0];
            x = x.substr(1);

            if (operator == ')') {
                return a;
            }

            var b = primary();
            a = (operator == '+') ? a + b :
                (operator == '-') ? a - b :
                (operator == '*') ? a * b :
                                    a / b;
        }
    }

    return expression();
}

这两个版本都不适用于IE，因为它们在字符串上使用数组样式的下标。如果您将x[0]的两个匹配项替换为x.charAt(0)，则第一个应该可以在任何地方使用。

我通过将变量转换为函数参数并用条件运算符替换另一个if语句，从第一个版本开始删除了一些字符。

Answer 12

C＃Regex Love

字符数： 384

<强>全混淆：

float E(string i){i=i.Replace(" ","");Regex b=new Regex(@"\((?>[^()]+|\((?<D>)|\)(?<-D>))*(?(D)(?!))\)");i=b.Replace(i,m=>Eval(m.Value.Substring(1,m.Length-2)).ToString());float r=0;foreach(Match m in Regex.Matches(i,@"(?<=^|\D)-?[\d.]+")){float f=float.Parse(m.Value);if(m.Index==0)r=f;else{char o=i[m.Index-1];if(o=='+')r+=f;if(o=='-')r-=f;if(o=='*')r*=f;if(o=='/')r/=f;}}return r;}

不会混淆的：

private static float Eval(string input) { input = input.Replace(" ", ""); Regex balancedMatcher = new Regex(@"$ (?> [^()]+ | \( (?<Depth>) | $ (?<-Depth>) )* (?(Depth)(?!)) \)", RegexOptions.IgnorePatternWhitespace); input = balancedMatcher.Replace(input, m => Eval(m.Value.Substring(1, m.Length - 2)).ToString()); float result = 0; foreach (Match m in Regex.Matches(input, @"(?<=^|\D)-?[\d.]+")) { float floatVal = float.Parse(m.Value); if (m.Index == 0) { result = floatVal; } else { char op = input[m.Index - 1]; if (op == '+') result += floatVal; if (op == '-') result -= floatVal; if (op == '*') result *= floatVal; if (op == '/') result /= floatVal; } } return result; }

利用.NET的正则表达式balancing group feature。

Answer 13

SQL（SQL Server 2008）

字符数：4202

完全混淆的功能：

WITH Input(id,str)AS(SELECT 1,'1 + 3 / -8'UNION ALL SELECT 2,'2*3*4*5+99'UNION ALL SELECT 3,'4 * (9 - 4)/ (2 * 6 - 2)+ 8'UNION ALL SELECT 4,'1 + ((123 * 3 - 69)/ 100)'UNION ALL SELECT 5,'2.45/8.5*9.27+(5*0.0023)'),Separators(i,ch,str_src,priority)AS(SELECT 1,'-',1,1UNION ALL SELECT 2,'+',1,1UNION ALL SELECT 3,'*',1,1UNION ALL SELECT 4,'/',1,1UNION ALL SELECT 5,'(',0,0UNION ALL SELECT 6,')',0,0),SeparatorsStrSrc(str,i)AS(SELECT CAST('['AS varchar(max)),0UNION ALL SELECT str+ch,SSS.i+1FROM SeparatorsStrSrc SSS INNER JOIN Separators S ON SSS.i=S.i-1WHERE str_src<>0),SeparatorsStr(str)AS(SELECT str+']'FROM SeparatorsStrSrc WHERE i=(SELECT COUNT(*)FROM Separators WHERE str_src<>0)),ExprElementsSrc(id,i,tmp,ele,pre_ch,input_str)AS(SELECT id,1,CAST(LEFT(str,1)AS varchar(max)),CAST(''AS varchar(max)),CAST(' 'AS char(1)),SUBSTRING(str,2,LEN(str))FROM Input UNION ALL SELECT id,CASE ele WHEN''THEN i ELSE i+1 END,CAST(CASE WHEN LEFT(input_str,1)=' 'THEN''WHEN tmp='-'THEN CASE WHEN pre_ch LIKE(SELECT str FROM SeparatorsStr)THEN tmp+LEFT(input_str,1)ELSE LEFT(input_str,1)END WHEN LEFT(input_str,1)IN(SELECT ch FROM Separators)OR tmp IN(SELECT ch FROM Separators)THEN LEFT(input_str,1)ELSE tmp+LEFT(input_str,1)END AS varchar(max)),CAST(CASE WHEN LEFT(input_str,1)=' 'THEN tmp WHEN LEFT(input_str,1)='-'THEN CASE WHEN tmp IN(SELECT ch FROM Separators)THEN tmp ELSE''END WHEN LEFT(input_str,1)IN(SELECT ch FROM Separators)OR tmp IN(SELECT ch FROM Separators)THEN CASE WHEN tmp='-'AND pre_ch LIKE(SELECT str FROM SeparatorsStr)THEN''ELSE tmp END ELSE''END AS varchar(max)),CAST(LEFT(ele,1)AS char(1)),SUBSTRING(input_str,2,LEN(input_str))FROM ExprElementsSrc WHERE input_str<>''OR tmp<>''),ExprElements(id,i,ele)AS(SELECT id,i,ele FROM ExprElementsSrc WHERE ele<>''),Scanner(id,i,val)AS(SELECT id,i,CAST(ele AS varchar(max))FROM ExprElements WHERE ele<>''UNION ALL SELECT id,MAX(i)+1,NULL FROM ExprElements GROUP BY id),Operator(op,priority)AS(SELECT ch,priority FROM Separators WHERE priority<>0),Calc(id,c,i,pop_count,s0,s1,s2,stack,status)AS(SELECT Scanner.id,1,1,0,CAST(scanner.val AS varchar(max)),CAST(NULL AS varchar(max)),CAST(NULL AS varchar(max)),CAST(''AS varchar(max)),CAST('init'AS varchar(max))FROM Scanner WHERE Scanner.i=1UNION ALL SELECT Calc.id,Calc.c+1,Calc.i,3,NULL,NULL,NULL,CASE Calc.s1 WHEN'+'THEN CAST(CAST(Calc.s2 AS real)+CAST(Calc.s0 AS real)AS varchar(max))WHEN'-'THEN CAST(CAST(Calc.s2 AS real)-CAST(Calc.s0 AS real)AS varchar(max))WHEN'*'THEN CAST(CAST(Calc.s2 AS real)*CAST(Calc.s0 AS real)AS varchar(max))WHEN'/'THEN CAST(CAST(Calc.s2 AS real)/CAST(Calc.s0 AS real)AS varchar(max))ELSE NULL END+' '+stack,CAST('calc '+Calc.s1 AS varchar(max))FROM Calc INNER JOIN Scanner NextVal ON Calc.id=NextVal.id AND Calc.i+1=NextVal.i WHERE Calc.pop_count=0AND ISNUMERIC(Calc.s2)=1AND Calc.s1 IN(SELECT op FROM Operator)AND ISNUMERIC(Calc.s0)=1AND(SELECT priority FROM Operator WHERE op=Calc.s1)>=COALESCE((SELECT priority FROM Operator WHERE op=NextVal.val),0)UNION ALL SELECT Calc.id,Calc.c+1,Calc.i,3,NULL,NULL,NULL,s1+' '+stack,CAST('paren'AS varchar(max))FROM Calc WHERE pop_count=0AND s2='('AND ISNUMERIC(s1)=1AND s0=')'UNION ALL SELECT Calc.id,Calc.c+1,Calc.i,Calc.pop_count-1,s1,s2,CASE WHEN LEN(stack)>0THEN SUBSTRING(stack,1,CHARINDEX(' ',stack)-1)ELSE NULL END,CASE WHEN LEN(stack)>0THEN SUBSTRING(stack,CHARINDEX(' ',stack)+1,LEN(stack))ELSE''END,CAST('pop'AS varchar(max))FROM Calc WHERE Calc.pop_count>0UNION ALL SELECT Calc.id,Calc.c+1,Calc.i+1,Calc.pop_count,CAST(NextVal.val AS varchar(max)),s0,s1,coalesce(s2,'')+' '+stack,cast('read'as varchar(max))FROM Calc INNER JOIN Scanner NextVal ON Calc.id=NextVal.id AND Calc.i+1=NextVal.i WHERE NextVal.val IS NOT NULL AND Calc.pop_count=0AND((Calc.s0 IS NULL OR calc.s1 IS NULL OR calc.s2 IS NULL)OR NOT(ISNUMERIC(Calc.s2)=1AND Calc.s1 IN(SELECT op FROM Operator)AND ISNUMERIC(calc.s0)=1AND (SELECT priority FROM Operator WHERE op=Calc.s1)>=COALESCE((SELECT priority FROM Operator WHERE op=NextVal.val),0))AND NOT(s2='('AND ISNUMERIC(s1)=1AND s0=')')))SELECT Calc.id,Input.str,Calc.s0 AS result FROM Calc INNER JOIN Input ON Calc.id=Input.id WHERE Calc.c=(SELECT MAX(c)FROM Calc calc2 WHERE Calc.id=Calc2.id)ORDER BY id

清除/半混淆功能：

WITH
  Input(id, str) AS (    
    SELECT 1, '1 + 3 / -8'
    UNION ALL SELECT 2, '2*3*4*5+99'
    UNION ALL SELECT 3, '4 * (9 - 4) / (2 * 6 - 2) + 8'
    UNION ALL SELECT 4, '1 + ((123 * 3 - 69) / 100)'
    UNION ALL SELECT 5, '2.45/8.5*9.27+(5*0.0023)'
  )
, Separators(i, ch, str_src, priority) AS (
    SELECT 1, '-', 1, 1
    UNION ALL SELECT 2, '+', 1, 1
    UNION ALL SELECT 3, '*', 1, 1
    UNION ALL SELECT 4, '/', 1, 1
    UNION ALL SELECT 5, '(', 0, 0
    UNION ALL SELECT 6, ')', 0, 0
  )
, SeparatorsStrSrc(str, i) AS (
    SELECT CAST('[' AS varchar(max)), 0
    UNION ALL
    SELECT
        str + ch
      , SSS.i + 1
    FROM
        SeparatorsStrSrc SSS
          INNER JOIN Separators S ON SSS.i = S.i - 1
    WHERE
        str_src <> 0
  )
, SeparatorsStr(str) AS (
    SELECT str + ']' FROM SeparatorsStrSrc
    WHERE i = (SELECT COUNT(*) FROM Separators WHERE str_src <> 0)
  )
, ExprElementsSrc(id, i, tmp, ele, pre_ch, input_str) AS (
    SELECT
        id
      , 1
      , CAST(LEFT(str, 1) AS varchar(max))
      , CAST('' AS varchar(max))
      , CAST(' ' AS char(1))
      , SUBSTRING(str, 2, LEN(str))
    FROM
        Input
    UNION ALL
    SELECT
        id
      , CASE ele
        WHEN '' THEN i
                ELSE i + 1
        END
      , CAST(
          CASE
          WHEN LEFT(input_str, 1) = ' '
            THEN ''
          WHEN tmp = '-'
            THEN CASE
                 WHEN pre_ch LIKE (SELECT str FROM SeparatorsStr)
                   THEN tmp + LEFT(input_str, 1)
                   ELSE LEFT(input_str, 1)
                 END
          WHEN LEFT(input_str, 1) IN (SELECT ch FROM Separators)
               OR
               tmp IN (SELECT ch FROM Separators)
            THEN LEFT(input_str, 1)
            ELSE tmp + LEFT(input_str, 1)
          END
        AS varchar(max))
      , CAST(
          CASE
          WHEN LEFT(input_str, 1) = ' '
            THEN tmp
          WHEN LEFT(input_str, 1) = '-'
            THEN CASE
                 WHEN tmp IN (SELECT ch FROM Separators)
                   THEN tmp
                   ELSE ''
                 END
          WHEN LEFT(input_str, 1) IN (SELECT ch FROM Separators)
               OR
               tmp IN (SELECT ch FROM Separators)
            THEN CASE
                 WHEN tmp = '-' AND pre_ch LIKE (SELECT str FROM SeparatorsStr)
                   THEN ''
                   ELSE tmp
                 END
            ELSE ''
          END
        AS varchar(max))
      , CAST(LEFT(ele, 1) AS char(1))
      , SUBSTRING(input_str, 2, LEN(input_str))
    FROM
        ExprElementsSrc
    WHERE
        input_str <> ''
        OR
        tmp <> ''
  )
, ExprElements(id, i, ele) AS (
    SELECT
        id
      , i
      , ele
    FROM
        ExprElementsSrc
    WHERE
        ele <> ''
  )
, Scanner(id, i, val) AS (
    SELECT
        id
      , i
      , CAST(ele AS varchar(max))
    FROM
        ExprElements
    WHERE
        ele <> ''
    UNION ALL
    SELECT
        id
      , MAX(i) + 1
      , NULL
    FROM
        ExprElements
    GROUP BY
        id
  )
, Operator(op, priority) AS (
    SELECT
        ch
      , priority 
    FROM
        Separators
    WHERE
        priority <> 0
  )
, Calc(id, c, i, pop_count, s0, s1, s2, stack, status) AS (
    SELECT
        Scanner.id
      , 1
      , 1
      , 0
      , CAST(scanner.val AS varchar(max))
      , CAST(NULL AS varchar(max))
      , CAST(NULL AS varchar(max))
      , CAST('' AS varchar(max))
      , CAST('init' AS varchar(max))
    FROM
        Scanner
    WHERE
        Scanner.i = 1
    UNION ALL
    SELECT
        Calc.id
      , Calc.c + 1
      , Calc.i
      , 3
      , NULL
      , NULL
      , NULL
      , CASE Calc.s1
        WHEN '+' THEN CAST(CAST(Calc.s2 AS real) + CAST(Calc.s0 AS real) AS varchar(max))
        WHEN '-' THEN CAST(CAST(Calc.s2 AS real) - CAST(Calc.s0 AS real) AS varchar(max))
        WHEN '*' THEN CAST(CAST(Calc.s2 AS real) * CAST(Calc.s0 AS real) AS varchar(max))
        WHEN '/' THEN CAST(CAST(Calc.s2 AS real) / CAST(Calc.s0 AS real) AS varchar(max))
                 ELSE NULL
        END
          + ' '
          + stack
      , CAST('calc ' + Calc.s1 AS varchar(max))
    FROM
        Calc
          INNER JOIN Scanner NextVal ON Calc.id = NextVal.id
                                          AND Calc.i + 1 = NextVal.i
    WHERE
        Calc.pop_count = 0
          AND ISNUMERIC(Calc.s2) = 1
          AND Calc.s1 IN (SELECT op FROM Operator)
          AND ISNUMERIC(Calc.s0) = 1
          AND (SELECT priority FROM Operator WHERE op = Calc.s1)
            >= COALESCE((SELECT priority FROM Operator WHERE op = NextVal.val), 0)
    UNION ALL
    SELECT
        Calc.id
      , Calc.c + 1
      , Calc.i
      , 3
      , NULL
      , NULL
      , NULL
      , s1 + ' ' + stack
      , CAST('paren' AS varchar(max))
    FROM
        Calc
    WHERE
        pop_count = 0
          AND s2 = '('
          AND ISNUMERIC(s1) = 1
          AND s0 = ')'
    UNION ALL
    SELECT
        Calc.id
      , Calc.c + 1
      , Calc.i
      , Calc.pop_count - 1
      , s1
      , s2
      , CASE
        WHEN LEN(stack) > 0
          THEN SUBSTRING(stack, 1, CHARINDEX(' ', stack) - 1)
          ELSE NULL
        END
      , CASE
        WHEN LEN(stack) > 0
          THEN SUBSTRING(stack, CHARINDEX(' ', stack) + 1, LEN(stack))
          ELSE ''
        END
      , CAST('pop' AS varchar(max))
    FROM
        Calc
    WHERE
        Calc.pop_count > 0
    UNION ALL
    SELECT
        Calc.id
      , Calc.c + 1
      , Calc.i + 1
      , Calc.pop_count
      , CAST(NextVal.val AS varchar(max))
      , s0
      , s1
      , coalesce(s2, '') + ' ' + stack
      , cast('read' as varchar(max))
    FROM
        Calc
          INNER JOIN Scanner NextVal ON Calc.id = NextVal.id
                                          AND Calc.i + 1 = NextVal.i
    WHERE
        NextVal.val IS NOT NULL
          AND Calc.pop_count = 0
          AND (
            (Calc.s0 IS NULL or calc.s1 is null or calc.s2 is null)
            OR
            NOT(
              ISNUMERIC(Calc.s2) = 1
                AND Calc.s1 IN (SELECT op FROM Operator)
                AND ISNUMERIC(calc.s0) = 1
                AND (SELECT priority FROM Operator WHERE op = Calc.s1)
                  >= COALESCE((SELECT priority FROM Operator WHERE op = NextVal.val), 0)
            )
              AND NOT(s2 = '(' AND ISNUMERIC(s1) = 1 AND s0 = ')')
          )
  )
SELECT
    Calc.id
  , Input.str
  , Calc.s0 AS result
FROM
    Calc
      INNER JOIN Input ON Calc.id = Input.id
WHERE
    Calc.c = (SELECT MAX(c) FROM Calc calc2
              WHERE Calc.id = Calc2.id)
ORDER BY
    id

它不是最短的。但我认为它对SQL非常灵活。添加新运算符很容易。改变运营商的优先级很容易。

Answer 14

PHP

字符数：284

混淆：

function f($m){return c($m[1]);}function g($n,$m){$o=$m[0];$m[0]=' ';return$o=='+'?$n+$m:($o=='-'?$n-$m:($o=='*'?$n*$m:$n/$m));}function c($s){while($s!=($t=preg_replace_callback('/\(([^()]*)\)/',f,$s)))$s=$t;preg_match_all('![-+/*].*?[\d.]+!',"+$s",$m);return array_reduce($m[0],g);}

可读：

function callback1($m) {return c($m[1]);}
function callback2($n,$m) {
    $o=$m[0];
    $m[0]=' ';
    return $o=='+' ? $n+$m : ($o=='-' ? $n-$m : ($o=='*' ? $n*$m : $n/$m));
}
function c($s){ 
    while ($s != ($t = preg_replace_callback('/\(([^()]*)\)/','callback1',$s))) $s=$t;
    preg_match_all('![-+/*].*?[\d.]+!', "+$s", $m);
    return array_reduce($m[0], 'callback2');
}


$str = '  2.45/8.5  *  -9.27   +    (   5   *  0.0023  ) ';
var_dump(c($str));
# float(-2.66044117647)

应该使用任何有效输入（包括负数和任意空格）

Answer 15

J

字符数：208

在Jeff Moser评论之后，我意识到我完全忘记了这种语言......我不是专家，但我的第一次尝试进展顺利。

e=:>@{:@f@;:
f=:''&(4 :0)
'y x'=.x g y
while.($y)*-.')'={.>{.y do.'y x'=.(x,>(-.'/'={.>{.y){('%';y))g}.y end.y;x
)
g=:4 :0
z=.>{.y
if.z='('do.'y z'=.f}.y else.if.z='-'do.z=.'_',>{.}.y end.end.(}.y);":".x,z
)

有点烦人，不得不将x/y和-z映射到J的x%y和_z。没有它，可能有50％的代码可能会消失。

Answer 16

F＃

字符数：327

OP正在寻找一个F＃版本，就在这里。因为我在这里滥用 ref 来保存字符，所以可以做得更好。它处理大多数事情，如 - （1.0）， 3 - -3 甚至 0 - .5 等。

let g s=
 let c=ref[for x in System.Text.RegularExpressions.Regex.Matches(s,"[0-9.]+|[^\s]")->x.Value]
 let rec e v=if (!c).IsEmpty then v else 
  let h=(!c).Head
  c:=(!c).Tail
  match h with|"("->e(e 0.0)|")"->v|"+"->e(v+(e 0.0))|"-"->e(v-(e 0.0))|"/"->e(v/(e 0.0))|"*"->e(v*(e 0.0))|x->float x
 e(e 0.0)

Answer 17

Python（不导入任何东西）

字符数：222

我从戴夫的回答中偷走了许多技巧，但我还是设法减少了一些角色。

def e(s,l=0,n=0,f='+'):
 if s:l=[c for c in s+')'if' '!=c]
 while f!=')':
  p=l.pop;m=p(0)
  if m=='(':m=e(0,l)
  while l[0]not in'+-*/)':m+=p(0)
  m=float(m);n={'+':n+m,'-':n-m,'*':n*m,'/':n/(m or 1)}[f];f=p(0)
 return n

评论版：

def evaluate(stringexpr, listexpr=0, n=0, f_operation='+'):
    # start out as taking 0 + the expression... (or could use 1 * ;)

    # We'll prefer to keep the expression as a list of characters,
    # so we can use .pop(0) to eat up the expression as we go.
    if stringexpr:
        listexpr = [c for c in stringexpr+')' if c!=' ']

    # use ')' as sentinel to return the answer
    while f_operation != ')':
        m_next = listexpr.pop(0)
        if m_next == '(':
            # lists are passed by reference, so this call will eat the (parexp)
            m_next = evaluate(None, listexpr)

        else:
            # rebuild any upcoming numeric chars into a string
            while listexpr[0] not in '+-*/)':
                m_next += listexpr.pop(0)

        # Update n as the current answer.  But never divide by 0.
        m = float(m_next)
        n = {'+':n+m, '-':n-m, '*':n*m, '/':n/(m or 1)}[f_operation]

        # prepare the next operation (known to be one of '+-*/)')
        f_operation = listexpr.pop(0)

    return n

Answer 18

MATLAB（v7.8.0）

字符数：239

混淆功能：

function [v,s]=m(s),r=1;while s,s=regexp(s,'( ?)(?(1)-?)[\.\d]+|\S','match');c=s{end};s=[s{1:end-1}];if any(c>47),v=str2num(c);elseif c>41,[l,s]=m(s);v=[l/v l*v l+v l-v];v=v(c=='/*+-');if r,break;end;r=1;elseif c<41,break;end;r=r&c~=41;end

清除（呃）功能：

function [value,str] = math(str)
  returnNow = 1;
  while str,
    str = regexp(str,'( ?)(?(1)-?)[\.\d]+|\S','match');
    current = str{end};
    str = [str{1:end-1}];
    if any(current > 47),
      value = str2num(current);
    elseif current > 41,
      [leftValue,str] = math(str);
      value = [leftValue/value leftValue*value ...
               leftValue+value leftValue-value];
      value = value(current == '/*+-');
      if returnNow,
        break;
      end;
      returnNow = 1;
    elseif current < 41,
      break;
    end;
    returnNow = returnNow & (c ~= 41);
  end

测试：

>> [math('1 + 3 / -8'); ...
math('2*3*4*5+99'); ...
math('4 * (9 - 4) / (2 * 6 - 2) + 8'); ...
math('1 + ((123 * 3 - 69) / 100)'); ...
math('2.45/8.5*9.27+(5*0.0023)')]

ans =

   -0.5000
  219.0000
   10.0000
    4.0000
    2.6834

概要：正则表达式和递归的混合。几乎是我迄今为止所做的最好的事情，没有作弊和使用EVAL。

Answer 19

这是另一个：

Shell脚本（使用sed + awk）

字符数：295

混淆：

e(){ a="$1";while echo "$a"|grep -q \(;do eval "`echo "$a"|sed 's/\(.*\)(\([^()]*\))\(.*\)/a="\1\`e \"\2\"\`\3"/'`";done; echo "$a"|sed 's/\([-+*/]\) *\(-\?\) */ \1 \2/g'|awk '{t=$1;for(i=2;i<NF;i+=2){j=$(i+1);if($i=="+") t+=j; else if($i=="-") t-=j; else if($i=="*") t*=j; else t/=j}print t}';}

可读

e () {
    a="$1"
    # Recursively process bracket-expressions
    while echo "$a"|grep -q \(; do
        eval "`echo "$a"|
            sed 's/\(.*\)(\([^()]*\))\(.*\)/a="\1\`e \"\2\"\`\3"/'`"
    done
    # Compute expression without brackets
    echo "$a"|
        sed 's/\([-+*/]\) *\(-\?\) */ \1 \2/g'|
        awk '{
            t=$1;
            for(i=2;i<NF;i+=2){
                j=$(i+1);
                if($i=="+") t+=j;
                else if($i=="-") t-=j;
                else if($i=="*") t*=j;
                else t/=j
            }
            print t
        }'
}

测试：

str='  2.45 / 8.5  *  9.27   +    (   5   *  0.0023  ) '
echo "$str"|bc -l
e "$str"

结果：

2.68344117647058823526
2.68344

Answer 20

C＃

字符数：403

所以这是我的解决方案......我还在等待有人在C＃中发布一个可以击败它的人。（Marc Gravell很接近，经过一些修补后可能会比我更好。）

完全混淆的功能：

float e(string x){float v=0;if(float.TryParse(x,out v))return v;x+=';';int t=0;
char o,s='?',p='+';float n=0;int l=0;for(int i=0;i<x.Length;i++){o=s;if(
x[i]!=' '){s=x[i];if(char.IsDigit(x[i])|s=='.'|(s=='-'&o!='1'))s='1';if(s==')')
l--;if(s!=o&l==0){if(o=='1'|o==')'){n=e(x.Substring(t,i-t));if(p=='+')v+=n;
if(p=='-')v-=n;if(p=='*')v*=n;if(p=='/')v/=n;p=x[i];}t=i;if(s=='(')t++;}
if(s=='(')l++;}}return v;}

半混淆功能：

public static float Eval(string expr)
{
    float val = 0;
    if (float.TryParse(expr, out val))
        return val;
    expr += ';';
    int tokenStart = 0;
    char oldState, state = '?', op = '+';
    float num = 0;
    int level = 0;
    for (int i = 0; i < expr.Length; i++)
    {
        oldState = state;
        if (expr[i] != ' ')
        {
            state = expr[i];
            if (char.IsDigit(expr[i]) || state == '.' ||
                (state == '-' && oldState != '1'))
                state = '1';
            if (state == ')')
                level--;
            if (state != oldState && level == 0)
            {
                if (oldState == '1' || oldState == ')')
                {
                    num = Eval(expr.Substring(tokenStart, i - tokenStart));
                    if (op == '+') val += num;
                    if (op == '-') val -= num;
                    if (op == '*') val *= num;
                    if (op == '/') val /= num;
                    op = expr[i];
                }
                tokenStart = i;
                if (state == '(')
                    tokenStart++;
            }
            if (state == '(')
                level++;
        }
    }
    return val;
}

没有什么比这更聪明了，看起来好像。但是，该函数具有重入（即线程安全）的优点。

我也很满意字符的数量，因为它是用C＃编写的（我相信有效的1.0,2.0和3.0）。

Answer 21

红宝石

字符数：170

混淆：

def s(x)
while x.sub!(/\(([^\(\)]*?)\)/){s($1)}
x.gsub!('--','')
end
while x.sub!(/(-?[\d.]+)[ ]*([+\-*\/])[ ]*(-?[\d.]+)/){$1.to_f.send($2,$3.to_f)}
end
x.strip.to_f
end

可读：

def s(x)
while x.sub!(/\(([^\(\)]*?)\)/){s($1)}
x.gsub!('--','')
end
while x.sub!(/(-?[\d.]+)[ ]*([+\-*\/])[ ]*(-?[\d.]+)/){$1.to_f.send($2,$3.to_f)}
end
x.strip.to_f
end

[
  ['1 + 3 / -8', -0.5],
  ['2*3*4*5+99', 219],
  ['4 * (9 - 4) / (2 * 6 - 2) + 8', 10],
  ['1 + ((123 * 3 - 69) / 100)', 4],
  ['2.45/8.5*9.27+(5*0.0023)',2.68344117647059],
  ['(3+7) - (5+2)', 3]
].each do |pair|
  a,b = s(String.new(pair[0])),pair[1]
  print pair[0].ljust(25), ' = ', b, ' (', a==b, ')'
  puts
end

对于这一个没有真正的混淆，我决定发布新鲜的，因为它与我的第一个完全不同。我应该从一开始就看到这一点。该过程是一个非常简单的消除过程：找到并解析最高的一对括号（最嵌套的）到一个数字，直到找不到更多，然后将所有现有的数字和操作解析为结果。并且，在解析括号语句时，我将它删除所有双破折号（Float.to_f不知道如何处理它们）。

因此，它支持正数和负数（+ 3,3和-3），甚至只是按处理顺序否定括号内的子表达式。唯一较短的实现是Perl（w / o eval）。

编辑：我还在追逐Perl，但这是目前第二小的答案。我通过更改第二个正则表达式并将字符串的处理更改为破坏性（替换旧字符串）来缩小它。这消除了复制字符串的需要，我发现它只是一个新的字符串指针。并将函数重命名为 s 从解决中保存了几个字符。

Answer 22

的Python

字符数：235

完全混淆的功能：

def g(a):
 i=len(a)
 while i:
  try:m=g(a[i+1:]);n=g(a[:i]);a=str({'+':n+m,'-':n-m,'*':n*m,'/':n/(m or 1)}[a[i]])
  except:i-=1;j=a.rfind('(')+1
  if j:k=a.find(')',j);a=a[:j-1]+str(g(a[j:k]))+a[k+1:]
 return float(a.replace('--',''))

半混淆：

def g(a):
    i=len(a);
    # do the math
    while i:
        try:
            # recursively evaluate left and right
            m=g(a[i+1:])
            n=g(a[:i])
            # try to do the math assuming that a[i] is an operator
            a=str({'+':n+m,'-':n-m,'*':n*m,'/':n/(m or 1)}[a[i]])
        except:
            # failure -> next try
            i-=1
            j=a.rfind('(')+1
        # replace brackets in parallel (this part is executed first)
        if j:
            k=a.find(')',j)
            a=a[:j-1]+str(g(a[j:k]))+a[k+1:]
    return float(a.replace('--',''))

FWIW，第n + 1个Python解决方案。在公然滥用尝试 - 除了我使用试错法。它应该正确处理所有情况，包括-(8)，--8和g('-(1 - 3)')等内容。它是可重入的。如果不支持许多实现不支持的--情况，则为217个字符（参见上一版本）。

感谢周日有趣的一个小时，周一再过30分钟。感谢krubo他的好词。

Answer 23

红宝石

字符数： ~~217~~ 179

~~这是迄今为止最短的ruby解决方案（当字符串包含几组括号时，一个基于RegExp的错误答案）~~ - 不再是真的。基于正则表达式和替换的解决方案更短。这个基于累加器堆栈并从左到右解析整个表达式。它是可重入的，不会修改输入字符串。它可能会被指责违反不使用eval的规则，因为它将Float的方法称为与其数学助记符（+， - ，/，*）相同的名称。

混淆代码（旧版本，在下面调整）：

def f(p);a,o=[0],['+']
p.sub(/-/,'+-').scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each{|n|
q,w=n;case w;when'(';a<<0;o<<'+';when')';q=a.pop;else;o<<w
end if q.nil?;a[-1]=a[-1].method(o.pop).call(q.to_f) if !q.nil?};a[0];end

更混淆的代码：

def f(p);a,o=[0],[:+]
p.scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each{|n|q,w=n;case w
when'(';a<<0;o<<:+;when')';q=a.pop;else;o<<w;end if !q
a<<a.pop.send(o.pop,q.to_f)if q};a[0];end

清洁代码：

def f(p)
  accumulators, operands = [0], ['+']
  p.gsub(/-/,'+-').scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each do |n|
    number, operand = n
    case operand
      when '('
        accumulators << 0
        operands << '+'
      when ')'
        number = accumulators.pop
        operands.pop
      else 
        operands[-1] = operand
    end if number.nil?
    accumulators[-1] = accumulators.last.method(operands[-1]).call(number.to_f) unless number.nil?
  end
  accumulators.first
end

Answer 24

C＃

字符数：396 （已更新）

（但是你用“/ -8”添加的测试失败了，我不打算修复它......

static float Eval(string s){int i,j;s=s.Trim();while((i=s.IndexOf(')'))>=0){j=s.LastIndexOf('(',i,i);s=s.Substring(0,j++)+Eval(s.Substring(j,i-j))+s.Substring(i+1);}if((i=s.LastIndexOfAny("+-*/".ToCharArray()))<0) return float.Parse(s);var r=float.Parse(s.Substring(i+1));var l=i>0?Eval(s.Substring(0,i)):(float?)null;return s[i]=='+'?(l??0)+r:(s[i]=='-'?(l??0)-r:(s[i]=='/'?(l??1)/r:(l??1)*r));}

自：

static float Eval(string s)
{
    int i, j;
    s = s.Trim();
    while ((i = s.IndexOf(')')) >= 0)
    {
        j = s.LastIndexOf('(', i, i);
        s = s.Substring(0, j++) + Eval(s.Substring(j, i - j)) + s.Substring(i + 1);
    } 
    if ((i = s.LastIndexOfAny("+-*/".ToCharArray())) < 0) return float.Parse(s);
    var r = float.Parse(s.Substring(i + 1));
    var l = i > 0 ? Eval(s.Substring(0, i)) : (float?)null;
    return s[i] == '+'
        ? (l ?? 0) + r
        : (s[i] == '-'
            ? (l ?? 0) - r
            : (s[i] == '/'
                ? (l ?? 1) / r
                : (l ?? 1) * r));
}

Answer 25

的Python

字符数：382

另一种Python解决方案，大量使用正则表达式替换。每次运行循环时，计算最简单的表达式，并将结果放回到字符串中。

这是未经模糊处理的代码，除非您考虑对正则表达式进行模糊处理。

import re
from operator import *    
operators = dict(zip("+-/*", (add, sub, truediv, mul)))    
def compute(s):
    def repl(m):
        v1, op, v2 = m.groups()
        return str(operators[op](float(v1), float(v2)))
    while not re.match("^\d+\.\d+$", s):
        s = re.sub("([.\d]+)\s*([+-/*])\s*([.\d]+)", repl, s)
        s = re.sub("\(([.\d]+)\)", r"\1", s)
    return s

这个想法就像我上交时一样，在我把它写下来并让它发挥作用之前不能放过它。

Answer 26

带有正则表达式的Python

字符数：283

完全混淆的功能：

import re
from operator import*
def c(e):
 O=dict(zip("+-/*()",(add,sub,truediv,mul)))
 a=[add,0];s=a
 for v,o in re.findall("(-?[.\d]+)|([+-/*()])",e):
  if v:s=[float(v)]+s
  elif o=="(":s=a+s
  elif o!=")":s=[O[o]]+s
  if v or o==")":s[:3]=[s[1](s[2],s[0])]
 return s[0]

没有混淆：

import re
from operator import *

def compute(s):
    operators = dict(zip("+-/*()", (add, sub, truediv, mul)))
    stack = [add, 0]
    for val, op in re.findall("(-?[.\d]+)|([+-/*()])", s):
        if val:
            stack = [float(val)] + stack
        elif op == "(":
            stack = [add, 0] + stack
        elif op != ")":
            stack = [operators[op]] + stack
        if val or op == ")":
            stack[:3] = [stack[1](stack[2], stack[0])]
    return stack[0]

我想看看我是否使用正则表达式击败其他Python解决方案。

不能。

我正在使用的正则表达式创建了一对（val，op）列表，其中每对中只有一个项有效。其余的代码是一个相当标准的基于堆栈的解析器，其中有一个巧妙的技巧，即使用Python列表赋值语法替换堆栈中的前3个单元格。使用负数进行此操作只需要两个附加字符（ - 正则表达式中的 - ？）。

Answer 27

SNOBOL4

字符数：232

        a = pos(0) | '('
        n = span('0123456789.')
        j = '!+;!-;!*;!/;       output = e'
d       j '!' len(1) . y = "    e a . q n . l '" y "' n . r = q (l " y " r)     :s(p)"  :s(d)
        k = code(j)
        e = input
s       e ' ' = :s(s)
p       e ('(' n . i ')') = i   :s(p)f<k>
end

这是一个半骗子。它使用code()（eval的变体）来解压缩自身，但不评估输入表达式。

去模糊版本，没有code：

        prefix = pos(0) | '('
        num = span('0123456789.')
        expr = input
spaces  expr ' ' = ''   :s(spaces)
paren   expr ('(' num . x ')') = x      :s(paren)
add     expr (prefix . pfx) (num . l) '+' (num . r) = pfx (l + r)       :s(paren)
sub     expr (prefix . pfx) (num . l) '-' (num . r) = pfx (l - r)       :s(paren)
mul     expr (prefix . pfx) (num . l) '*' (num . r) = pfx (l * r)       :s(paren)
div     expr (prefix . pfx) (num . l) '/' (num . r) = pfx (l / r)       :s(paren)
        output = expr
end

策略：

首先，删除所有空格（spaces）
尽可能删除数字（paren）
否则，找一个涉及两个数字的简单表达式，前缀为'('或字符串的开头
如果上述规则均不适用，则表达式将被完全评估。现在，如果输入结构良好，我们应该留下一个数字。

示例：

1 + (2 * 3) + 4
1+(2*3)+4 [spaces]
1+(6)+4 [mul]
1+6+4 [paren]
7+4 [add]
11 [add]

Answer 28

C＃with Regex

字符数： 294

这部分取决于Jeff Moser的回答，但采用了极为简化的评估技术。甚至可能还有其他方法可以减少字数，但我现在很高兴有一个300字符以下的C＃解决方案！

完全混淆代码：

float e(string x){while(x.Contains("("))x=Regex.Replace(x,@"\(([^\(]*?)\)",m=>e(m.Groups[1].Value).ToString());float r=0;foreach(Match m in Regex.Matches("+"+x,@"\D ?-?[\d.]+")){var o=m.Value[0];var v=float.Parse(m.Value.Substring(1));r=o=='+'?r+v:o=='-'?r-v:o=='*'?r*v:r/v;}return r;}

更清晰的代码：

float e(string x)
{
    while (x.Contains("("))
        x = Regex.Replace(x, @"\(([^\(]*?)\)", m => e(m.Groups[1].Value).ToString());
    float r = 0;
    foreach (Match m in Regex.Matches("+" + x, @"\D ?-?[\d.]+"))
    {
        var o = m.Value[0];
        var v = float.Parse(m.Value.Substring(1));
        r = o == '+' ? r + v : o == '-' ? r - v : o == '*' ? r * v : r / v;
    }
    return r;
}

Answer 29

Ruby 1.9

（因为正则表达式）

字符数：296

def d(s)
  while m = s.match(/((?<pg>\((?:\\[()]|[^()]|\g<pg>)*\)))/)
    s.sub!(m[:pg], d(m[:pg][1,m[:pg].size-2]))
  end
  while m = s.match(/(-?\d+(\.\d+)?)\s*([*+\-\/])\s*(-?\d+(\.\d+)?)/)
    r=m[1].to_f.send(m[3],m[4].to_f) if %w{+ - * /}.include?m[3]
    s.sub!(m[0], r.to_s)
  end
  s
end

编辑：包括马丁的优化。

Answer 30

C＃

字符数：355

我拿了Noldorin's Answer并修改了它，所以给予Noldorin 99％的功劳。我能用算法做的最好的就是408个字符。有关更清晰的代码版本，请参阅Noldorin's Answer。

所做的更改：
更改字符比较以与数字进行比较删除了一些默认声明并组合了相同类型的声明重新处理了一些if语句。

float q(string x){float v,n;if(!float.TryParse(x,out v)){x+=';';int t=0,l=0,i=0;char o,s='?',p='+';for(;i<x.Length;i++){o=s;if(x[i]!=32){s=x[i];if(char.IsDigit(x[i])|s==46|(s==45&o!=49))s='1';if(s==41)l--;if(s!=o&l==0){if(o==49|o==41){n=q(x.Substring(t,i-t));v=p==43?v+n:p==45?v-n:p==42?v*n:p==47?v/n:v;p=x[i];}t=i;if(s==40)t++;}if(s==40)l++;}}}return v;}

编辑：通过删除其中一个返回声明，将其从361减少到355。

Answer 31

Ruby 1.8.7

字符数：620

尝试在我的实现上轻松实现，这是我第一次在我的生活中编写表达式解析器！我保证这不是最好的。

<强>混淆：

def solve_expression(e)
t,r,s,c,n=e.chars.to_a,[],'','',''
while(c=t.shift)
n=t[0]
if (s+c).match(/^(-?)[.\d]+$/) || (!n.nil? && n.match(/\d/) && c=='-')
s+=c
elsif (c=='-' && n=='(') || c=='('
m,o,x=c=='-',1,''
while(c=t.shift)
o+=1 if c=='('
o-=1 if c==')'
x+=c unless c==')' && o==0
break if o==0
end
r.push(m ? -solve_expression(x) : solve_expression(x))
s=''
elsif c.match(/[+\-\/*]/)
r.push(c) and s=''
else
r.push(s) if !s.empty?
s=''
end
end
r.push(s) unless s.empty?
i=1
a=r[0].to_f
while i<r.count
b,c=r[i..i+1]
c=c.to_f
case b
when '+': a=a+c
when '-': a=a-c
when '*': a=a*c
when '/': a=a/c
end
i+=2
end
a
end

<强>可读：

def solve_expression(expr)
  chars = expr.chars.to_a # characters of the expression
  parts = [] # resulting parts
  s,c,n = '','','' # current string, character, next character

  while(c = chars.shift)
    n = chars[0]
    if (s + c).match(/^(-?)[.\d]+$/) || (!n.nil? && n.match(/\d/) && c == '-') # only concatenate when it is part of a valid number
      s += c
    elsif (c == '-' && n == '(') || c == '(' # begin a sub-expression
      negate = c == '-'
      open = 1
      subExpr = ''
      while(c = chars.shift)
        open += 1 if c == '('
        open -= 1 if c == ')'
        # if the number of open parenthesis equals 0, we've run to the end of the
        # expression.  Make a new expression with the new string, and add it to the
        # stack.
        subExpr += c unless c == ')' && open == 0
        break if open == 0
      end
      parts.push(negate ? -solve_expression(subExpr) : solve_expression(subExpr))
      s = ''
    elsif c.match(/[+\-\/*]/)
      parts.push(c) and s = ''
    else
      parts.push(s) if !s.empty?
      s = ''
    end
  end
  parts.push(s) unless s.empty? # expression exits 1 character too soon.

  # now for some solutions!
  i = 1
  a = parts[0].to_f # left-most value is will become the result
  while i < parts.count
    b,c = parts[i..i+1]
    c = c.to_f
    case b
      when '+': a = a + c
      when '-': a = a - c
      when '*': a = a * c
      when '/': a = a / c
    end
    i += 2
  end
  a
end

Answer 32

红宝石

字符数： 302

<强>半混淆：

def e(l)
  t=0.0;o=nil
  while l!=''
    l.sub!(/^\s+/,'')
    l.sub!(/^(-?\d+|-?\d+\.\d+)/,'')
    t=o ? t.send(o, $1.to_f) : $1.to_f if $~
    l.sub!(/^(\+|-|\*|\/)/,'')
    o=$1 if $~
    l.sub!(/^\(/,'')
    t=o ? t.send(o, e(l)) : e(l) if $~
    l.sub!(/^\)/,'')
    return t if $~
  end
  t
end

销毁原始字符串，也假设表达式格式正确（只有有效字符和匹配括号）。

未混淆：

def evaluate_expression(expression)
  result_so_far = 0.0
  last_operator = nil

  while (expression != '')
    # remove any leading whitespace
    expression.sub!(/^\s+/, '') 

    # extract and remove leading integer or decimal number
    expression.sub!(/^(-?\d+|-?\d+\.\d+)/, '')
    if $~
      # match was successful
      number = $1.to_f
      if last_operator.nil?
        # first number, just store it
        result_so_far = number
      else
        # we have an operator, use it!
        # last_operator is a string matching '+', '-', '*' or '/'
        # just invoke the method of that name on our result_so_far
        # since these operators are just method calls in Ruby
        result_so_far = result_so_far.send(last_operator, number)
       end
    end

    # extract and remove leading operator +-*/
    expression.sub!(/^(\+|-|\*|\/)/, '')
    if $~
      # match was successful
      last_operator = $1
    end

    # extract and remove leading open bracket
    l.sub!(/^\(/, '')
    if $~
      # match successful
      if last_operator.nil?
        # first element in the expression is an open bracket
        # so just evaluate its contents recursively
        result_so_far = evaluate_expression(expression)
      else
        # combine the content of the bracketing with the
        # result so far using the last_operator
        result_so_far.send(last_operator, evaluate_expression(expression))
      end
    end

    # extract and remove leading close bracket
    l.sub!(/^\)/, '')
    if $~
      # match successful
      # this must be the end of a recursive call so
      # return the result so far without consuming the rest
      # of the expression
      return result_so_far
    end
  end
  t
end

递归调用是由表达式字符串的修改控制的，这有点令人讨厌，但似乎有用。

Answer 33

Python 3K

（它的3K因为/将结果转换为浮点数）

字符数：808

清除（我不能在Python XD中编写混淆代码）：

def parse(line):
  ops = {"+": lambda x,y:x+y,
       "-": lambda x,y:x-y,
       "*": lambda x,y:x*y,
       "/": lambda x,y:x/y}
  def tpp(s, t):
    if len(s) > 0 and s[-1] in ops:
      f = ops[s.pop()]
      t = f(s.pop(), t)
    return t
  line = line + " "
  s = []
  t = 0
  m = None
  for c in line:
    if c in "0123456789":
      if not m:
        m = "i"
      if m == "i":
        t = t*10 + ord(c)-ord("0")
      elif m =="d":
        t = t + e*(ord(c)-ord("0"))
        e*=0.1
    elif c == ".":
      m = "d"
      e = 0.1
    elif m:
      t = tpp(s,t)
      s.append(t)
      m = None
      t = 0

    if c in ops or c == "(":
      s.append(c)
    elif c == ")":
      t = s.pop()
      s.pop()
      s.append(tpp(s,t))
      t = 0
  t = s.pop()
  if int(t) == t:
    t = int(t)
  return t

我没有使用任何类型的正则表达式，即使数字解析是手工制作的; - ）

非常简单，扫描线，它可以是3种不同的模式（m），None表示没有被解析的数字，“i”表示它正在解析整数部分而“d”表示这意味着正在解析小数部分。

它使用一个堆栈来存储临时计算，当它完成解析一个数字时，看看堆栈中是否有一个运算符，在这种情况下是evals和push。开口的parens被推开，关闭的parens移除开口paren并重新压缩当前的eval。

相当简单和直接： - ）

Answer 34

的Python

字符数：492

模糊混淆的函数（短变量名，操作符周围没有空格）：

def e(s):
    q=[]
    b=1
    v=[]
    for c in s.replace(' ','')+'$':
        if c in '.0123456789' or c in '+-' and b and not v:
            v+=[c]
        else:
            if v:
                q+=[float(''.join(v))]
                v=[]
            while len(q)>=3:
                x,y,z=q[-3:]
                if type(x)==type(z)==float:
                    if y=='+':q[-3:]=[x+z]
                    elif y=='-':q[-3:]=[x-z]
                    elif y=='*':q[-3:]=[x*z]
                    elif y=='/':q[-3:]=[x/z]
                elif (x,z)==('(',')'):q[-3:]=[y]
                else:break
            if c=='$':break
            q+=[c]
            b=c!=')'
    return q[0]

我认为这相对容易理解。这是一种非常直接，天真的方法。它不导入任何东西，不使用正则表达式，完全自包含（单个函数，没有全局，没有副作用），并且应该处理有符号的文字（正面或负面）。使用更合理的变量名称并遵循推荐的Python格式将字符数增加到850-900，这是使用四个空格而不是单个制表符进行缩进的大部分内容。

Answer 35

的Java

字符数：376

更新版本，现在有更多？操作员滥用！

完全混淆的解决方案：

static double e(String t){t="("+t+")";for(String s:new String[]{"+","-","*","/","(",")"})t=t.replace(s," "+s+" ");return f(new Scanner(t));}static double f(Scanner s){s.next();double a,v=s.hasNextDouble()?s.nextDouble():f(s);while(s.hasNext("[^)]")){char o=s.next().charAt(0);a=s.hasNextDouble()?s.nextDouble():f(s);v=o=='+'?v+a:o=='-'?v-a:o=='*'?v*a:v/a;}s.next();return v;}

清除/半混淆功能：

static double evaluate(String text) {
    text = "(" + text + ")";
    for (String s : new String[] {"+", "-", "*", "/", "(", ")" }) {
        text = text.replace(s, " " + s + " ");
    }
    return innerEval(new Scanner(text));
}

static double innerEval(Scanner s) {
    s.next();
    double arg, val = s.hasNextDouble() ? s.nextDouble() : innerEval(s);
    while (s.hasNext("[^)]")) {
        char op = s.next().charAt(0);
        arg = s.hasNextDouble() ? s.nextDouble() : innerEval(s);
        val =
            op == '+' ? val + arg :
            op == '-' ? val - arg :
            op == '*' ? val * arg :
            val / arg;
    }
    s.next();
    return val;
}

Answer 36

F＃

字符数：461

这是Marc Gravell的解决方案（基本上）从C＃转换为F＃。焦炭数量显然更好，但我认为无论如何我都会发布它的兴趣。

混淆代码：

let e x=
 let rec f(s:string)=
  let i=s.IndexOf(')')
  if i>0 then
   let j=s.LastIndexOf('(',i)
   f(s.Substring(0,j)+f(s.Substring(j+1,i-j-1))+s.Substring(i+1))
  else
   let o=[|'+';'-';'*';'/'|]
   let i=s.LastIndexOfAny(o)
   let j=s.IndexOfAny(o,max(i-2)0,2)
   let k=if j<0 then i else j
   if k<0 then s else
    let o=s.[k]
    string((if o='+'then(+)else if o='-'then(-)else if o='*'then(*)else(/))(float(f(s.Substring(0,k))))(float(s.Substring(k+1))))
 float(f x)

Answer 37

C ++

Chars：1670

 // not trying to be terse here
#define DIGIT(c)((c)>='0' && (c) <= '9')
#define WHITE(pc) while(*pc == ' ') pc++
#define LP '('
#define RP ')'

bool SeeNum(const char* &pc, float& fNum){
    WHITE(pc);
    if (!(DIGIT(*pc) || (*pc=='.'&& DIGIT(pc[1])))) return false;
    const char* pc0 = pc;
    while(DIGIT(*pc)) pc++;
    if (*pc == '.'){
        pc++;
        while(DIGIT(*pc)) pc++;
    }
    char buf[200];
    int len = pc - pc0;
    strncpy(buf, pc0, len); buf[len] = 0;
    fNum = atof(buf);
    return true;
}

bool SeeChar(const char* &pc, char c){
    WHITE(pc);
    if (*pc != c) return false;
    pc++;
    return true;
}

void ParsExpr(const char* &pc, float &fNum);

void ParsPrim(const char* &pc, float &fNum){
    if (SeeNum(pc, fNum));
    else if (SeeChar(pc, LP)){
        ParsExpr(pc, fNum);
        if (!SeeChar(pc, RP)) exit(0);
    }
    else exit(0); // you can abort better than this
}

void ParsUnary(const char* &pc, float &fNum){
    if (SeeChar(pc, '-')){
        pc+;
        ParsUnary(pc, fNum);
        fNum = -fNum;
    }
    else {
        ParsPrim(pc, fNum);
    }
}

void ParsExpr(const char* &pc, float &fNum){
    ParsUnary(pc, fNum);
    float f1 = 0;
    while(true){
        if (SeeChar(pc, '+')){
            ParsUnary(pc, f1);
            fNum += f1;
        }
        else if (SeeChar(pc, '-')){
            ParsUnary(pc, f1);
            fNum -= f1;
        }
        else if (SeeChar(pc, '*')){
            ParsUnary(pc, f1);
            fNum *= f1;
        }
        else if (SeeChar(pc, '/')){
            ParsUnary(pc, f1);
            fNum /= f1;
        }
        else break;
    }
}

这只是LL1（递归下降）。我喜欢这样做（虽然我使用双打），因为它足够快，并且易于插入例程来处理优先级。

Answer 38

的 PowerBASIC

字符数： ~400

有点难看，但它有效。 :)我确信regexp会让它更小。

DEFDBL E,f,i,z,q,a,v,o DEFSTR s,c,k,p FUNCTION E(s) i=LEN(s) DO IF MID$(s,i,1)="("THEN q=INSTR(i,s,")") s=LEFT$(s,i-1)+STR$(E(MID$(s,i+1,q-i-1)))+MID$(s,q+1) END IF i-=1 LOOP UNTIL i=0 k="+-*/" DIM p(PARSECOUNT(s,ANY k)) PARSE s,p(),ANY k a=VAL(p(0)) FOR i=1TO LEN(s) c=MID$(s,i,1) q=INSTR(k,c) IF q THEN z+=1 IF o=0 THEN o=q ELSE p(z)=c+p(z) IF TRIM$(p(z))<>"" THEN v=VAL(p(z)) a=CHOOSE(o,a+v,a-v,a*v,a/v) o=0 END IF END IF NEXT E=a END FUNCTION

Answer 39

C＃，264个字符

策略：前两行通过归纳摆脱括号。然后我按\-?[\d.]+分割得到数字和运算符。然后使用aggregate将字符串数组减少为double值。

可变解释

m是带括号的表达式，没有嵌套括号。
d是这种笨拙的TryParse语法的占位符。
v是最终值的累加器
t是当前令牌。

float E(string s){var d=999f;while(d-->1)s=Regex.Replace(s,@"(([^(]?))",m=>E(m.Groups[1].Value)+"");return Regex.Split(s,@"(-?[\d.]+)").Aggregate(d,(v,t)=>(t=t.Trim()).Length==0?v:!float.TryParse(t,out d)?(s=t)==""?0:v:s=="/"?v/d:s=="-"?v-d:s==""?v*d:v+d);}

    float F(string s) {
        var d=999f;
        while(d-->1)
            s=Regex.Replace(s,@"\(([^\(]*?)\)",m=>F(m.Groups[1].Value)+"");
        return Regex.Split(s, @"(\-?[\d\.]+)")
            .Aggregate(d, (v, t) => 
                (t=t.Trim()).Length == 0 ? v :
                !float.TryParse(t, out d) ? (s=t) == "" ? 0 : v :
                s == "/" ? v / d :
                s == "-" ? v - d :
                s == "*" ? v * d :
                           v + d);
    }

编辑：无耻地从noldorin的答案中偷走了部分，重新使用s作为操作员变量。

编辑：999嵌套括号对任何人都应该足够了。

Answer 40

OCaml using Camlp4 directly：

open Camlp4.PreCast

let expr = Gram.Entry.mk "expr"

EXTEND Gram
  expr:
  [   [ e1 = expr; "+"; e2 = expr -> e1 + e2
      | e1 = expr; "-"; e2 = expr -> e1 - e2 ]
  |   [ e1 = expr; "*"; e2 = expr -> e1 * e2
      | e1 = expr; "/"; e2 = expr -> e1 / e2 ]
  |   [ n = INT -> int_of_string n
      | "("; e = expr; ")" -> e ]   ];
END

let () = Gram.parse expr Loc.ghost (Stream.of_string "1-2+3*4")

OCaml using the Camlp4 stream parser extension：

open Genlex

let lex = make_lexer ["+"; "-"; "*"; "/"; "("; ")"]

let rec parse_atom = parser
  | [< 'Int n >] -> n
  | [< 'Kwd "("; e=parse_expr; 'Kwd ")" >] -> e
and parse_factor = parser
  | [< e1=parse_atom; stream >] ->
      (parser
         | [< 'Kwd "*"; e2=parse_factor >] -> e1 * e2
         | [< 'Kwd "/"; e2=parse_factor >] -> e1 / e2
         | [< >] -> e1) stream
and parse_expr = parser
  | [< e1=parse_factor; stream >] ->
      (parser
         | [< 'Kwd "+"; e2=parse_expr >] -> e1 + e2
         | [< 'Kwd "-"; e2=parse_expr >] -> e1 - e2
         | [< >] -> e1) stream

let () =
  Printf.printf "%d\n" (parse_expr(lex(Stream.of_string "1 + 2 * (3 + 4)")));;

Answer 41

我很惊讶没有人在Lex / Yacc或同等版本中做到这一点。

这似乎会产生最易读/可维护的源代码。

Answer 42

PHP

字符数：170

完全混淆的功能：

function a($a,$c='#\(([^()]*)\)#e',$d='a("$1","#^ *-?[\d.]+ *\S *-?[\d.]+ *#e","\$0")'){$e='preg_replace';while($a!=$b=$e($c,$d,$a))$a = $b;return$e('#^(.*)$#e',$d,$a);}

更清晰的功能：

function a($a, $c = '#\(([^()]*)\)#e', $d = 'a("$1", "#^ *-?[\d.]+ *\S *-?[\d.]+ *#e", "\$0")') {
    $e = 'preg_replace';
    while ($a != $b = $e($c, $d, $a)) {
        $a = $b;
    }
    return $e('#^(.*)$#e', $d, $a);
}

试验：

assert(a('1 + 3 / -8') === '-0.5');
assert(a('2*3*4*5+99') === '219');
assert(a('4 * (9 - 4) / (2 * 6 - 2) + 8') === '10');
assert(a('1 + ((123 * 3 - 69) / 100)') === '4');
assert(a('2.45/8.5*9.27+(5*0.0023)') === '2.68344117647');
assert(a(' 2 * 3 * 4 * 5 + 99 ') === '219');

Answer 43

的Perl

字符数：93

完全混淆的功能：（如果将这三行连接成一行，则为93个字符）

$_="(@ARGV)";s/\s//g;$n=qr/(-?\d+(\.\d+)?)/;
while(s.\($n\)|(?<=\()$n[-+*/]$n.eval$&.e){}
print

清除/半混淆功能：

$_="(@ARGV)";            # Set the default var to "(" argument ")"
s/\s//g;                 # Strip all spaces from $_
$n=qr/(-?\d+(\.\d+)?)/;  # Compile a regex for "number"

# repeatedly replace the sequence "(" NUM ")" with NUM, or if there aren't
# any of those, replace "(" NUM OP NUM with the result
# of doing an eval on just the NUM OP NUM bit.
while(s{\($n\)|(?<=\()$n[-+*/]$n}{eval$&}e){}

# print $_
print

我认为在“清晰”版本中可以很好地解释这一点。两个主要的见解是，您可以通过在开头用括号括起参数来使代码统一（特殊情况成本字符），并且仅仅处理开放式括号旁边的东西就足够了，尽管效率非常低，替换它的结果。

运行此代码可能最简单：

perl -le '$_="(@ARGV)";s/\s//g;$n=qr/(-?\d+(\.\d+)?)/;while(s.\($n\)|(?<=\()$n[-+*/]$n.eval$&.e){}print' '4 * (9 - 4) / (2 * 6 - 2) + 8'

评估一串简单的数学表达式

挑战

规则

我的解决方案

更新

格式

语言

43 个答案:

汇编

Perl（无评估）

更短

更短

从石头中挤水

C（VS2005）

红宝石

的Haskell

Visual Basic.NET

的Python

Fortran 77（gfortran方言，现在支持g77）

C99

Common Lisp

JavaScript（不兼容IE）

C＃Regex Love

SQL（SQL Server 2008）

字符数：4202

PHP

J

字符数：208

F＃

Python（不导入任何东西）

MATLAB（v7.8.0）

Shell脚本（使用sed + awk）

C＃

红宝石

的Python

红宝石

C＃

的Python

带有正则表达式的Python

SNOBOL4

C＃with Regex

Ruby 1.9

Ruby 1.8.7

红宝石

Python 3K

的Python

的Java

F＃

C ++

的 PowerBASIC

C＃，264个字符

PHP

的Perl