cl:编译时间与运行时间:inf与-nan(ind) [英] cl: compile time vs. run time: inf vs. -nan(ind)

问题描述

示例代码( t50.c ):

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
#include <math.h>
#include <float.h>
#include <assert.h>

const float d1 = NAN;
const float d2 = -0x0p+0;
const float d3 = NAN / -0x0p+0;

typedef union { uint32_t u; float d; } u_t;

int main(void)
{
        u_t u1;
        u_t u2;
        u_t u3;

        u1.d = *(volatile float*)&d1 / *(volatile float*)&d2;
        u2.d = d3;
        u3.d = d1 / d2;
        if ( u1.u != u2.u || u1.u != u3.u )
        {
                printf("error:\n");
                printf("u1 (run time)     %08"PRIx32" %.*e\n", u1.u, DECIMAL_DIG, u1.d);
                printf("u2 (compile time) %08"PRIx32" %.*e\n", u2.u, DECIMAL_DIG, u2.d);
                printf("u3                %08"PRIx32" %.*e\n", u3.u, DECIMAL_DIG, u3.d);
        }
        return 0;
}

编译器调用: cl t50.c/O1/fp:precise&&t50

预期结果:< nothing>

实际结果(与 cl x86 和 cl x64 相同):

Actual result (same for cl x86 and cl x64):

error:
u1 (run time)     ffc00000 -nan(ind)
u2 (compile time) 7f800000 inf
u3                ffc00000 -nan(ind)

我指定了/fp:strict : cl t50.c/O1/fp:strict&&t50 ，但得到了:

I've specified /fp:strict: cl t50.c /O1 /fp:strict && t50, but got:

t50.c(8): error C2099: initializer is not a constant
t50.c(10): error C2099: initializer is not a constant

cl 版本:x86的 19.25.28611 和x64的 19.25.28611 .

cl version: 19.25.28611 for x86 and 19.25.28611 for x64.

与 gcc (10.2.0)和 clang (11.0.0)进行比较:

Compare with gcc (10.2.0) and clang (11.0.0):

gcc t50.c -O2 && a.exe
<nothing>

clang t50.c -O2 && a.exe
<nothing>

为什么?我在这里想念的是什么?标准(C/IEEE)怎么说?

Why? What I'm missing here? What the standard (C / IEEE) says?

UPD1:

1. if(u1.d！= u2.d || u1.d！= u3.d)的结果相同.
2. 相同的结果不包含联合(即使用 float u1，u2，u3 ).

1. Same result for if ( u1.d != u2.d || u1.d != u3.d ).
2. Same result w/o unions (i.e. using float u1, u2, u3).

UPD2:

1. 编译器的NAN定义: Windows Kits \ 10 \ Include \ 10.0.18362.0 \ ucrt \ corecrt_math.h : #define NAN((float)(INFINITY * 0.0F)).
2. u1.d = NAN的输出printf("NAN％08" PRIx32％.* e \ n"，u1.u，DECIMAL_DIG，u1.d); (对于 cl x86 和 clx64 ): NAN ffc00000 -nan(ind).

1. compiler's NAN definition: Windows Kits\10\Include\10.0.18362.0\ucrt\corecrt_math.h: #define NAN ((float)(INFINITY * 0.0F)).
2. Output for u1.d = NAN; printf("NAN %08"PRIx32" %.*e\n", u1.u, DECIMAL_DIG, u1.d); (for both cl x86 and cl x64): NAN ffc00000 -nan(ind).

推荐答案

为什么?

cl 不遵循关于 IEEE-754 > NAN .

cl does not follow IEEE-754 concerning NAN.

我在这里想念什么?

What I'm missing here?

假设遵循IEEE标准的C编译器.

Assuming a compliant C compiler follows IEEE.

标准C怎么说?

C对 NAN 部门的规范不严格.我认为编译器是合规的，只是关于 NAN 的实现质量较差.

C is lax with specs in the NAN department. I'd rate the compiler as compliant, just a weak quality of implementation concerning NAN.

IEEE的标准是什么?

What the standard IEEE says?

不符合规定.

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