Why does C print my hex values incorrectly?

Question

So I'm a bit of a newbie to C and I am curious to figure out why I am getting this unusual behavior.

I am reading a file 16 bits at a time and just printing them out as follows.

#include <stdio.h>

#define endian(hex) (((hex & 0x00ff) << 8) + ((hex & 0xff00) >> 8))

int main(int argc, char *argv[])
 {
  const int SIZE = 2;
  const int NMEMB = 1;
  FILE *ifp; //input file pointe
  FILE *ofp; // output file pointer

  int i;
  short hex;
  for (i = 2; i < argc; i++)
   {
    // Reads the header and stores the bits
    ifp = fopen(argv[i], "r");
    if (!ifp) return 1;
    while (fread(&hex, SIZE, NMEMB, ifp))
     {
      printf("\n%x", hex);
      printf("\n%x", endian(hex)); // this prints what I expect
      printf("\n%x", hex);
      hex = endian(hex);
      printf("\n%x", hex);
     }   
   }
 }

The results look something like this:

ffffdeca
cade // expected
ffffdeca
ffffcade
0
0 // expected
0
0
600
6 // expected
600
6

Can anyone explain to me why the last line in each block doesn't print the same value as the second?

Answer 1

The placeholder %x in the format string interprets the corresponding parameter as unsigned int.

To print the parameter as short, add a length modifier h to the placeholder:

printf("%hx", hex);

http://en.wikipedia.org/wiki/Printf_format_string#Format_placeholders