我正在开发C ++的程序。我必须使用数学库的 POW 执行一些计算。
当运行此代码时：
#include <math.h>
#include <stdio.h>
int main (void)
{
unsigned long l;
double d1,d2,d3;
l = 13831305406817642647;
d1 = 2.;
d2 = *((double*)(&l));
d3 = pow(2.,d2);
printf ("%lu,%20.20f\n",*((unsigned long *)(&d1)),d1);
printf ("%lu,%20.20f\n",*((unsigned long *)(&d2)),d2);
printf ("%lu,%20.20f\n",*((unsigned long *)(&d3)),d3);
return 0;
}
使用Fedora 35，我将获得：
4611686018427387904,2.00000000000000000000
13831305406817642647,-1.16674465427289475450
4601695688420081959,0.44542528011426657519
但是使用Fedora 36：
4611686018427387904,2.00000000000000000000
13831305406817642647,-1.16674465427289475450
4601695688420081958,0.44542528011426651968
我将双打转换为无符号的二进制图像，长时间检测任何更改。
在第三行中，您可以观察到这种变化。
当我用POW计算相同的值时，为什么会获得不同的结果？
有关Fedora版本的完整信息（使用 UNAME -R 获得）：
对于Fedora 35：5.17.11-200.fc35.x86_64
for Fedora 36：5.17.0-0-0.rc7.116。 FC36.x86_64
欢迎任何帮助。
I'm developing a program in C++. I have to perform some calculations with pow of math library.
When a run this code:
#include <math.h>
#include <stdio.h>
int main (void)
{
unsigned long l;
double d1,d2,d3;
l = 13831305406817642647;
d1 = 2.;
d2 = *((double*)(&l));
d3 = pow(2.,d2);
printf ("%lu,%20.20f\n",*((unsigned long *)(&d1)),d1);
printf ("%lu,%20.20f\n",*((unsigned long *)(&d2)),d2);
printf ("%lu,%20.20f\n",*((unsigned long *)(&d3)),d3);
return 0;
}
With Fedora 35 I obtain:
4611686018427387904,2.00000000000000000000
13831305406817642647,-1.16674465427289475450
4601695688420081959,0.44542528011426657519
But with Fedora 36:
4611686018427387904,2.00000000000000000000
13831305406817642647,-1.16674465427289475450
4601695688420081958,0.44542528011426651968
I converted the doubles to a binary image of unsigned long to detect any change for just one bit.
In the third line you can observe this kind of change.
Why I obtain a different result when I calculate over the same values with pow?
The full information about the version of Fedora (obtained with uname -r):
For Fedora 35: 5.17.11-200.fc35.x86_64
For Fedora 36: 5.17.0-0.rc7.116.fc36.x86_64
Any help will be welcome.
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评论（1）
我使用 X86-64 GCC 12.1 在编译器资源管理器中两次编译您的代码：。唯一的区别是优化水平，这导致输出有所不同。
在测试中，
-O1启用编译器在编译过程中计算d3并通过460169568420081958直接到printf（printf（printf）。有趣的是， x86-64 clang 能够用
-O1来计算d3，结果为4601695688420081959（与POW（2.，D2）相同。基本上，如果优化级别不是
-O0，则编译器将删除pow（）的使用，并且可行的恒定计算是可行的。我猜 gcc 使用自己的方法来计算pow（），而 clang call 调用pow> pow（）或使用获得答案的类似方法。您可以使用objdump从可执行文件中获取汇编代码以检查详细信息。I used x86-64 gcc 12.1 to compile your code twice in Compiler Explorer: https://godbolt.org/z/47K94dMe3. The only difference was the optimization level, which caused the outputs to differ.
In the test,
-O1enabled the compiler to calculate thed3during compilation and pass4601695688420081958directly toprintf().Interestingly, x86-64 clang is able to compute
d3with-O1, and the result is4601695688420081959(the same aspow(2.,d2)).Basically, compilers will remove the use of
pow()if optimization level is not-O0, and a constant calculation is viable. I guess gcc uses its own method to computepow(), while clang callspow()or uses a similar approach to get the answer. You can useobjdumpto get assembly code from your executables to check the details.