C - 添加两个单精度浮点正常数，不能得到无穷大的结果

Question

我正在玩浮点运算，我遇到了一些需要解释的东西。

将舍入模式设置为 'towards zero' 时，又名：

fesetround(FE_TOWARDZERO);

加上不同种类的正常正数，我永远无法达到无穷大。

但是，从 ieee 745 中得知，有限数相加可能导致溢出到无穷大。

例如：

#include <fenv.h>
#include <stdio.h>

float hex2float (int hex_num) {
  return *(float*)&hex_num;
}

void main() {
  int a_int = 0x7f7fffff; // Maximum finite single precision number, about 3.4E38
  int b_int = 0x7f7fffff;
  float a = hex2float(a_int);
  float b = hex2float(b_int);
  float res_add;

  fesetround(FE_TOWARDZERO);  // need to include fenv.h for that
  printf("Calculating... %+e + %+e\n",a,b);
  res_add = a + b;
  printf("Res = %+e\n",res_add);
}

但是，如果我将舍入模式更改为其他模式，我可能会得到 +INF 作为答案。

Can someone explain this?

Answer 1

对观察到的行为的解释是，它是 IEEE 754-2008 浮点标准强制要求的：

7.4 Overflow

The overflow exception shall be signaled if and only if the destination format’s largest finite number is exceeded in magnitude by what would have been the rounded floating-point result (see 4) were the exponent range unbounded. The default result shall be determined by the rounding-direction attribute and the sign of the intermediate result as follows:

[...]

b) roundTowardZero carries all overflows to the format’s largest finite number with the sign of the intermediate result.

所以对于这里使用的舍入模式（截断，或向零舍入），溢出情况下的结果是最大的有限数，不是无穷大。

C - 添加两个单精度浮点正常数，不能得到无穷大的结果

C - adding two single-precision floating point normal numbers, can't get result to infinity

c

floating-point

ieee-754

single-precision