定点类型不能正确相乘

时间:2013-06-02 20:26:55

标签: ada multiplication fixed-point

我是Ada的新手,并且一直在试用定点" delta"类型。具体来说,我创建了一个32位的delta类型范围0.0 .. 1.0。但是,当我尝试对某些值进行平方时,我得到一个CONSTRAINT_ERROR。据我所知,这应该发生在我指定的范围内。此错误的阈值似乎为sqrt(1/2)。我使用的是MinGW-w64版本4.8.0的GNAT。

测试代码(所有代码都以gnatmake <file>的形式编译,没有警告/错误):

types.ads:

pragma Ada_2012;

with Ada.Unchecked_Conversion;
with Ada.Text_IO;

package Types is
    type Fixed_Type is delta 1.0 / 2**32 range 0.0 .. 1.0
        with Size => 32;
    type Modular_Type is mod 2**32
        with Size => 32;
    function Fixed_To_Mod is new Ada.Unchecked_Conversion(Fixed_Type, Modular_Type);
    package MIO is new Ada.Text_IO.Modular_IO(Modular_Type);
    package FIO is new Ada.Text_IO.Fixed_IO(Fixed_Type);
end Types;

specifics.adb:

pragma Ada_2012;

with Ada.Text_IO;

with Types; use Types;

procedure Specifics is
    package TIO renames Ada.Text_IO;

    procedure TestValue(val: in Fixed_Type) is
        square : Fixed_Type;
    begin
        square := val * val;
        TIO.Put_Line("Value " & Fixed_Type'Image(val) & " squares properly.");
        TIO.Put_Line("Square: " & Fixed_Type'Image(square));
        TIO.New_Line;
    exception
        when Constraint_Error =>
            TIO.Put_Line("Value " & Fixed_Type'Image(val) & " does not square properly.");
            TIO.Put_Line("Square: " & Fixed_Type'Image(val * val));
            TIO.Put_Line("Not sure how that worked.");
            TIO.New_Line;
    end TestValue;

    function ParseFixed(s: in String; last: in Natural; val: out Fixed_Type) return Boolean is
        l : Natural;
    begin
        FIO.Get(s(s'First..last), val, l);
        return TRUE;
    exception
        when others =>
            TIO.Put_Line("Parsing failed.");
            return FALSE;
    end ParseFixed;

    buffer : String(1..20);
    last : Natural;
    f : Fixed_Type;
begin
    loop
        TIO.Put(">>> ");
        TIO.Get_Line(buffer, last);
        exit when buffer(1..last) = "quit";
        if ParseFixed(buffer, last, f) then
            TestValue(f);
        end if;
    end loop;
end Specifics;

specifics.adb的输出:

>>> 0.1
Value  0.1000000001 squares properly.
Square:  0.0100000000

>>> 0.2
Value  0.2000000000 squares properly.
Square:  0.0399999998

>>> 0.4
Value  0.3999999999 squares properly.
Square:  0.1599999999

>>> 0.6
Value  0.6000000001 squares properly.
Square:  0.3600000001

>>> 0.7
Value  0.7000000000 squares properly.
Square:  0.4899999998

>>> 0.75
Value  0.7500000000 does not square properly.
Square: -0.4375000000
Not sure how that worked.

>>> quit

不知何故,将val乘以它会产生一个负数,这解释了CONSTRAINT_ERROR ......但是没关系,为什么我首先得到一个负数?

然后我决定测试数字开始失败的平方点,所以我写了以下片段:

fixedpointtest.adb:

pragma Ada_2012;

with Ada.Text_IO;

with Types; use Types;

procedure FixedPointTest is
    package TIO renames Ada.Text_IO;

    test, square : Fixed_Type := 0.0;
begin
    while test /= Fixed_Type'Last loop
        square := test * test;
        test := test + Fixed_Type'Delta;
    end loop;
exception
    when Constraint_Error =>
        TIO.Put_Line("Last valid value: " & Fixed_Type'Image(test-Fixed_Type'Delta));
        TIO.Put("Hex value: ");
        MIO.Put(Item => Fixed_To_Mod(test-Fixed_Type'Delta), Base => 16);
        TIO.New_Line;
        TIO.Put("Binary value: ");
        MIO.Put(Item => Fixed_To_Mod(test-Fixed_Type'Delta), Base => 2);
        TIO.New_Line;
        TIO.New_Line;
        TIO.Put_Line("First invalid value: " & Fixed_Type'Image(test));
        TIO.Put("Hex value: ");
        MIO.Put(Item => Fixed_To_Mod(test), Base => 16);
        TIO.New_Line;
        TIO.Put("Binary value: ");
        MIO.Put(Item => Fixed_To_Mod(test), Base => 2);
        TIO.New_Line;
        TIO.New_Line;
end FixedPointTest;

并获得以下输出:

Last valid value:  0.7071067810
Hex value: 16#B504F333#
Binary value: 2#10110101000001001111001100110011#

First invalid value:  0.7071067812
Hex value: 16#B504F334#
Binary value: 2#10110101000001001111001100110100#

所以,sqrt(1/2),我们再次见面。有人可以向我解释为什么我的代码这样做?有没有办法让它正确繁殖?

1 个答案:

答案 0 :(得分:5)

我认为你要求的精确度比实际可用的“精确度”要低1倍。

您的声明

   type Fixed_Type is delta 1.0 / 2**32 range 0.0 .. 1.0
       with Size => 32;

仅被接受,因为GNAT使用了有偏见的表示;标志位没有空间。您可以看到这一点,因为0.7071067810表示为16#B504F333#,其中设置了最重要的位。因此,当您将0.71乘以0.71时,结果具有最高位设置;并且低级代码认为这必须是符号位,所以我们有溢出。

如果您将Fixed_Type声明为

   type Fixed_Type is delta 1.0 / 2**31 range 0.0 .. 1.0
       with Size => 32;
一切都应该好。

还有一点:在报告中specifics的行为输入为0.75时,引用结果

>>> 0.75
Value  0.7500000000 does not square properly.
Square: -0.4375000000
Not sure how that worked.

我使用gnatmake specifics.adb -g -gnato -bargs -E重建,结果现在是

>>> 0.75
Value  0.7500000000 does not square properly.

Execution terminated by unhandled exception
Exception name: CONSTRAINT_ERROR
Message: 64-bit arithmetic overflow
Call stack traceback locations:
0x100020b79 0x10000ea80 0x100003520 0x100003912 0x10000143e

并且回溯解码为

system__arith_64__raise_error (in specifics) (s-arit64.adb:364)
__gnat_mulv64 (in specifics) (s-arit64.adb:318)
specifics__testvalue.2581 (in specifics) (specifics.adb:20)        <<<<<<<<<<
_ada_specifics (in specifics) (specifics.adb:45)
main (in specifics) (b~specifics.adb:246)

specifics.adb:20

     TIO.Put_Line("Square: " & Fixed_Type'Image(val * val));

在异常处理程序中,它再次涉及有问题的方块(在异常处理程序中不是一件好事)。您可以看到上面一行中打印的值0.75没有任何问题:在fixedpointtest.adb中,添加中没有问题导致上一个有效值0.7071067810

我很惊讶地发现-gnato检测到这个错误,因为我认为它只适用于整数运算;但实际上在GNAT User Guide中有一个讨论,它说它也适用于定点算术。事实证明,您可以使用-gnato3

来避免约束错误并获得正确的算术结果
>>> 0.75
Value  0.7500000000 squares properly.
Square:  0.5625000000

但仅以使用任意多精度算术为代价 - 对于时间受限的系统来说不是一个好主意!