printf in dump routine

[Piet van Oostrum]

In article <652 at dsacg2.UUCP>, nor1675 at dsacg2 (Michael Figg) writes:
 `    .
 `    long  test_long;
 `    test_long = 1234;
 `    print_data((char *)&test_long,sizeof(test_long));
 `    .
 `    void print_data(pnt,num)
 `    char *pnt;
 `    int num;
 `    {
 `      int i;
 `      for(i = 0; i < num; i++)
 `      {
 `	 printf("%02x",*(pnt + i));
 `      }
 `    }

		 Any clues as to why I'm getting this output and how to do it
 `    right?  Thanks

The problem is that this code is highly unportable. So what you get depends
very much on the machine/compiler you have. There are two problems:

1. The ENDIAN-NESS of the machine. This means whether the bytes in a long
   are addressed from the most significant byte or from the least
   significant byte. What you are doing is taking a long and getting bytes
   out of it. On a BIG-ENDIAN machine (such as the 68000) the first byte
   you get is the upper (most significant) one, on a SMALL-ENDIAN (like the
   8086) it is the lower byte. This explains the difference between the
   Amiga and MS-DOS.

2. The char type that your compiler has. This can be signed or unsigned
   (old compilers usually have only signed. In your case both compilers
   apparently have signed chars, which means that a char that has its
   8th bit set is passed to printf as a negative integer, which cuases
   printf to print extra ff's (how many depends on the size of int, which
   -- not surprisingly -- is 4 bytes on the Amiga and 2 bytes on MS-DOS).
   This problem is easily solved by

	 printf("%02x",*(pnt + i) & 0xff);
or
	 printf("%02x",pnt[i] & 0xff);
-- 
Piet van Oostrum, Dept of Computer Science, University of Utrecht
Padualaan 14, P.O. Box 80.089, 3508 TB Utrecht, The Netherlands
Telephone: +31-30-531806. piet at cs.ruu.nl (mcvax!hp4nl!ruuinf!piet)