c++ - 如何在Intel ICC中启用long double

Question

我在MacOS上，并按照Using the Intel® Math Library指南实现第一个示例:

// real_math.c
#include <stdio.h> 
#include <mathimf.h>

int main() {
 float fp32bits;
 double fp64bits;
 long double fp80bits;
 long double pi_by_four = 3.141592653589793238/4.0;

// pi/4 radians is about 45 degrees
 fp32bits = (float) pi_by_four; // float approximation to pi/4
 fp64bits = (double) pi_by_four; // double approximation to pi/4
 fp80bits = pi_by_four; // long double (extended) approximation to pi/4

// The sin(pi/4) is known to be 1/sqrt(2) or approximately .7071067 
 printf("When x = %8.8f, sinf(x) = %8.8f \n", fp32bits, sinf(fp32bits));
 printf("When x = %16.16f, sin(x) = %16.16f \n", fp64bits, sin(fp64bits));
 printf("When x = %20.20Lf, sinl(x) = %20.20Lf \n", fp80bits, sinl(fp80bits));

 return 0; 
}

按照指南中的指示，我编译为:

icc real_math.c

执行时，我得到:

When x = 0.78539819, sinf(x) = 0.70710677 
When x = 0.7853981633974483, sin(x) = 0.7071067811865475 
When x = 0.78539816339744830952, sinl(x) = 0.00000000000000000000 

我已经进行了广泛的搜索，所有示例似乎都表明这应该是微不足道的。我想念什么？
我试图将-long_double传递给icc，但是没有任何变化。

Answer 1

好的，我知道了。此问题是由于省略1个字符L而导致的我的错误。具体而言，以某种方式在我自己的计算机上的代码中，我有:

 printf("When x = %20.20Lf, sinl(x) = %20.20f \n", fp80bits, sinl(fp80bits));

正确的代码是这样(注意%20.20f与%20.20Lf):

 printf("When x = %20.20Lf, sinl(x) = %20.20Lf \n", fp80bits, sinl(fp80bits));

我什至不知道这怎么可能发生，但这是根本原因。
谢谢大家的快速回答，并感谢@PeterCordes将我链接到 godbolt 。