Windows / msvc 上的 clang：为什么在 FE_UPWARD 下 printf("%.1f\n", 0.0) 打印 0.1 而不是 0.0？

Question

示例代码（t928.c）：

#include <stdio.h>
#include <fenv.h>
#if _MSC_VER && ! __clang__
#pragma fenv_access     (on)
#else
#pragma STDC    FENV_ACCESS ON
#endif

int main(void)
{
        int i = fesetround( FE_UPWARD );
        if ( ! i )
        {
                printf( "%.1f\n", 0.0 );
        }
        return 0;
}

调用：

$ clang t928.c -Wall -Wextra -std=c11 -ffp-model=strict -pedantic && ./a.exe
0.1

$ cl t928.c /std:c11 /Za /fp:strict && ./t928.exe
0.1

版本：

$ clang --version
clang version 12.0.0

$ cl
Microsoft (R) C/C++ Optimizing Compiler Version 19.28.29913 for x64

UPD20210824。用户 chux - 恢复 Monica 假设：

Some compilers have trouble properly including FP support when there is no FP activity in user code. Try adding some.

这是它：

int main(void)
{
        float f1 = 0.0f;
        float f2 = 0.0f;
        float f3 = 0.0f;
        int i = fesetround(FE_UPWARD);
        if ( ! i )
        {
                printf("%.1f\n", 0.0);
        }
        f3 = f1 + f2;
        printf("%.1f\n", f3);
        return 0;
}

调用：

$ gcc t928.c -Wall -Wextra -std=c11 -pedantic && ./a.exe
t928.c:6: warning: ignoring '#pragma STDC FENV_ACCESS' [-Wunknown-pragmas]
    6 | #pragma STDC    FENV_ACCESS     ON
      |
0.0
0.0

$ cl t928.c /std:c11 /Za /fp:strict && ./t928.exe
0.1
0.1

$ clang12 t928.c -Wall -Wextra -std=c11 -ffp-model=strict -pedantic && ./a.exe
0.1
0.1

UPD20210831。微软的回答：

This is an issue in the Universal CRT with our FE_UPWARD and FE_DOWNWARD rounding modes where some numbers will round up as though they had additional non-zero digits after them. This will be fixed in the Universal CRT and will be included in a future release of the Windows OS and Windows SDK.

Answer 1

因为是bug in the Universal CRT.

Windows / msvc 上的 clang：为什么在 FE_UPWARD 下 printf("%.1f\n", 0.0) 打印 0.1 而不是 0.0？

clang on Windows / msvc: why under FE_UPWARD printf("%.1f\n", 0.0) prints 0.1 instead of 0.0?

c

visual-c++

floating-point

clang

rounding