c++ - -Ofast 在使用 long double 时产生不正确的代码

Question

#include <cstdio>

int main(void)
{
    int val = 500;
    printf("%d\n", (int)((long double)val / 500));
    printf("%d\n", (int)((long double)500 / 500));
}

显然它应该输出1 1。但是如果用-Ofast编译，会输出0 1，为什么？

而如果你把500改成其他值(比如400)，用-Ofast编译，还是会输出 1 1。

带有 -Ofast 的编译器资源管理器:https://gcc.godbolt.org/z/YkX7fB

似乎这一行导致了问题。

Answer 1

-Ofast

Disregard strict standards compliance. -Ofast enables all -O3 optimizations. It also enables optimizations that are not valid for all standard-compliant programs. It turns on -ffast-math, -fallow-store-data-races and the Fortran-specific [...]

-ffast-math

Sets the options -fno-math-errno, -funsafe-math-optimizations, -ffinite-math-only, -fno-rounding-math, -fno-signaling-nans, -fcx-limited-range and -fexcess-precision=fast.

This option causes the preprocessor macro __FAST_MATH__ to be defined.

This option is not turned on by any -O option besides -Ofast since it can result in incorrect output for programs that depend on an exact implementation of IEEE or ISO rules/specifications for math functions. It may, however, yield faster code for programs that do not require the guarantees of these specifications.

结论:不要使用-ffast-math，除非你愿意得到像现在这样的惊喜。