How to enable long double in Intel ICC

Question

I'm on MacOS, and am following the guide Using the Intel® Math Library to implement the first example:

// 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; 
}

As directed in the guide, I compile with:

icc real_math.c

When I execute, I get:

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

I've searched wide and far, and all examples seem to show that this should be trivial. What am I missing?

I tried to pass -long_double to icc, but nothing changes.

Answer 1

Ok, I figured it out. This problem was caused by an error on my part due to the omission of 1 character: L. Specifically, somehow in my own code on my computer I had this:

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

and the correct code is this (note %20.20f vs %20.20Lf):

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

I'm not even sure how this could have happened, but this was the root cause.

Thank you everyone for your quick answers, and thanks @PeterCordes for linking me to godbolt.