Floating Point Arithmetic, C output

Question

for which value of f, the output will be "world"?

#include<stdio.h>
int main(){
  float f= ... ;
  if(f==f)
     printf("hello\n");
  else
     printf("world\n");
 return 0;
}

what will be replaced with three dots (...)

Answer 1

Try this:

#include<stdio.h>
#include<stdlib.h>
#include<math.h>

int main(){
  float f = nanf("0.0");
  printf("%f\n", f);

  if(f==f)
     printf("hello\n");
  else
     printf("world\n");
 return 0;
}

Answer 2

Easy.

float f = (puts("world"), exit(0), 0.0f);

Edit: Yes, I know you're looking for a different answer. But I'm going to be cute instead because this looks like a problem from a homework assignment or take-home quiz that hasn't even been broken down into its parts.

Answer 3

IEEE 754 floating point numbers can represent positive or negative infinity and NaN (not a number). These 3 values arise from calculations whose result is undefined or cannot be represented accurately. NaN will always propagate through expressions. ie any expression containing NaN will evaluate NaN .

You can also deliberately set a floating-point variable to NaN . For example, the following perfectly valid floating-point expression will do the trick :

 sqrtf(-1.0f) // NaN

More info on infinity and NaN in floating point numbers .

---

PS: I was tempted to ignore this as it looks like a straightforward homework Q. But what with the homework tag being deprecated and all...

Answer 4

You can use the NAN macro from math.h .

#include <math.h>
#include <stdio.h>

int main(void)
{
  float f = NAN;
  if (f == f)
     printf("hello\n");
  else
     printf("world\n");
}

Note that this macro has been introduced in C99 and is not present if the implementation does not support quiet NaNs for the float type.