我刚接触Ada，并尝试使用固定点“delta”类型。具体来说，我创建了一个32位的delta类型范围为0.0 .. 1.0。然而，当我尝试对某些值进行平方时，会出现CONSTRAINT_ERROR错误。据我所知，在我的指定范围内不应该发生这种情况。此错误的阈值似乎为
测试代码（所有代码都以
sqrt(1/2)。我正在使用MinGW-w64版本4.8.0的GNAT。测试代码（所有代码都以
gnatmake <file>形式编译，无警告/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乘以自身得到了一个负数，这解释了CONSTRAINT_ERROR...但是不管它，为什么一开始我就得到了一个负数呢？
然后我决定测试平方数字失败的点，所以我写了以下代码片段:
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)，我们再次见面了。有人能向我解释一下为什么我的代码会这样吗？有办法使其正确地进行乘法运算吗？
with Size => 32子句时才能编译通过；否则会出现编译错误）。对于原始的增量，我想知道为什么每当我尝试分配一个值 >= 0.5 时，程序没有引发CONSTRAINT_ERROR。然而，当我重新使用-gnato进行编译时，我发现它实际上不接受这样的值。是否有一种方法可以摆脱符号位，或者我必须使用不同的增量？ - ericmaht