Question

我修正了错误，但是现在可以有人建议让CountNearMatches计算两个数组中有多少匹配匹配的方法。我使用diff变量作为最大允许差异。

这是我到目前为止所做的：

    INCLUDE Irvine32.inc

CountNearMatches PROTO,           ; procedure prototype
    arr1:PTR SDWORD,
    arr2:PTR SDWORD,
    diff3:DWORD

.data

arraySyze DWORD 6
    diff DWORD 3
    index DWORD  0           ;var to use as the first index

    Array1  SDWORD  15, 9, 11, 13, 19, 6
    Array2  SDWORD  14, 5, 3, 12, 1, 4  

    text1 BYTE "************* ASM *****************", 0
    text2 BYTE "*** Counting Nearly Matching Elements ***", 0
    text3 BYTE "the arrays are:...", 0
    text4 BYTE "Nearly Mathces were: ",0

.code
main PROC

    mov edx, offset text1       ;hands the address of the firs char of text1
    call WriteString            ;prints out text1
    call Crlf
    call Crlf

    mov edx, offset text2       ; hands the address of the firs char of text2
    call WriteString            ; prints out text2
    call Crlf
    call Crlf

    mov edx, offset text3       ;hands the address of the firs char of text3
    call WriteString            ;prints out text3
    call Crlf

    mov  esi,OFFSET Array1      ; puts the offset of array in esi
    mov  ecx,arraySyze          ;6
    mov  ebx,TYPE Array1                            
    call DumpMem
    call Crlf

    mov  esi,OFFSET Array2      ; puts the offset of array in esi
    mov  ecx,arraySyze          ;  6
    mov  ebx,TYPE Array1                            
    call DumpMem

    mov eax,0
swapLoop:
    mov ebx, index              ;moves index1 to ebx


    ;here i use the registers to add 8 in to the stack addresses
    INVOKE CountNearMatches, ADDR[Array1 + ebx], ADDR[Array2 + ebx], diff

    add ebx, 4                  ;adding 4 to ebx
    mov index, ebx              ;putting ebx back in to index1

    loop swapLoop

    call Crlf

    mov edx, offset text4       ;hands the address of the firs char of text4
    call WriteString            ;prints out text4



    call WriteInt

    call Crlf
    call Crlf


    call WaitMsg                ; Press any key to contineu...
    exit
main ENDP

; ***************************************
; This is where the magic happnes!
; This function gets its parameters from
; the stack and counts the near mathches
; from the two arrays...
; ***************************************
CountNearMatches PROC USES ebx edx esi edi,
    arr1:PTR SDWORD,                    ;points to the first array
    arr2:PTR SDWORD,                    ;points to the second array
    diff3:DWORD                         ;gets the diff from the stack

;------------------------------------------------------- 

    mov eax,[arr1]
    mov ebx,[arr2]
    mov edx, diff3

    sub eax,ebx
    ret
CountNearMatches ENDP

END main

Answer 1

我想通了..现在它实际上计算了所有近距离比赛，不确定这是否是最好的方式，但它的工作原理。这就是我想出来的

var word = 'javascript'.split(''); // ['j', 'a', 'v', 'a' ...]

function guessLetter(guess) {
  if(word.indexOf(guess) > -1) {
    // remove correct letter guess from word
    word.splice(word.indexOf(guess), 1);
  }

  if(word.length === 0) {
    console.log('you win');
  }
}

guessLetter('j');

装配错误初始化程序幅度对于指定大小而言太大

1 个答案: