What is the result of casting float +INF, -INF, and NAN to integer in C?

Question

Does any standard specifies what should be the output?

For example this code:

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

int main(int argc, char** argv) {
  float a = INFINITY;
  float b = -INFINITY;
  float c = NAN;

  printf("float %f %f %f\n", a, b, c); 
  printf("int %d %d %d\n", (int) a, (int) b, (int) c); 
  printf("uint %u %u %u\n", (unsigned int) a, (unsigned int) b, (unsigned int) c); 
  printf("lint %ld %ld %ld\n", (long int) a, (long int) b, (long int) b); 
  printf("luint %lu %lu %lu\n", (unsigned long int) a, (unsigned long int) b, (unsigned long int) c); 

  return 0;
}

Compiled on gcc version 4.2.1 (Apple Inc. build 5664) Target: i686-apple-darwin10

Outputs:

$ gcc test.c && ./a.out 
float inf -inf nan
int -2147483648 -2147483648 -2147483648
uint 0 0 0
lint -9223372036854775808 -9223372036854775808 -9223372036854775808
luint 0 9223372036854775808 9223372036854775808

Which is quite weird. (int)+inf < 0 !?!

In Java infinities are cast to the closest value (MAX_VALUE or MIN_VALUE) and NaN is cast to 0. — starblue, Oct 21 '10 at 18:32

score 17 · Accepted Answer · answered Oct 21 '10 at 11:25

As Paul said, it's undefined:

From §6.3.1.4:

6.3.1.4 Real ﬂoating and integer

When a ﬁnite value of real ﬂoating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero). If the value of the integral part cannot be represented by the integer type, the behavior is undeﬁned.50)

Infinity isn't finite, and the integral part can't be represented in an integral type, so it's undefined.

What is the result of casting float +INF, -INF, and NAN to integer in C?

1 Answers1

Linked