This examples show how floating point numbers that are NaN are always different even if they are the same binary identity

nan.c

#include <stdio.h>

/* 
   malkia ~/p $ gcc nan.c
   nan.c: In function ‘print_range’:
   nan.c:15: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘unsigned int’
   nan.c:15: warning: format ‘%p’ expects type ‘void *’, but argument 3 has type ‘unsigned int’
   nan.c:15: warning: format ‘%p’ expects type ‘void *’, but argument 4 has type ‘unsigned int’
   malkia ~/p $ ./a.out 
   [0x7f800001 .. 0x7fffffff], total of 0x7fffff (8388607)
   [0xff800001 .. 0xffffffff], total of 0x7fffff (8388607)
*/

union {
  float f;
  unsigned u;
} n;

/*
  The following would make NaN not happy :)
  -ffast-math
  -ffinite-math-only
 */

void print_range( unsigned first, unsigned last )
{
  printf( "[%p .. %p], total of %p (%d)\n", first, last, last - first + 1, last - first + 1 );
}

int main()
{
  unsigned first_nan = 0, found_nan = 0;

  // 0 is not a NaN so we start from 1
  // And use the fact that it would wrap
  for( n.u=1; n.u; n.u++ ) {
    // Is it a NaN number?
    // A NaN number is never equal to itself
    if( n.f != n.f ) {
      if( !found_nan ) {
	found_nan = 1;
	first_nan = n.u;
      }
      continue;
    }
    if( found_nan ) {
      print_range( first_nan, n.u - 1 );
      found_nan = 0;
    }
  }
  if( found_nan ) {
    print_range( first_nan, n.u - 1 );
    found_nan = 0;
  }
  return 0;
}