INFO: Promotion of char to int May Add Two Hex Digits in printf

• Microsoft C for MS-DOS, versions 6.0, 6.0a, 6.0ax
• Microsoft C for OS/2, versions 6.0, 6.0a
• Microsoft C/C++ for MS-DOS, version 7.0
• Microsoft Visual C++ for Windows, versions 1.0, 1.5
• Microsoft Visual C++ 32-bit Edition, versions 1.0, 2.0, 4.0, 5.0

SUMMARY

When you try to display a signed character in two-digit hexadecimal format as follows:

   char hex_byte = 0x80;
   printf( "The Hex value is %X", hex_byte);

To preserve the two-digit display for all possible values, typecast the character to an unsigned character in the printf() argument list, as shown below:

   printf( "The hex value is %X", (unsigned char)hex_byte);

MORE INFORMATION

For ANSI compliance, the compiler promotes all character arguments to signed int. If the type is signed character, then the value will be sign- extended. Thus, the int will be negative if the left-most bit of the character was set, resulting in a different value for the int. When an unsigned character is promoted, a 0 (zero) is placed in the high byte so that the value is retained.

This promotion may also cause problems with comparisons of signed and unsigned characters. This situation, unlike the printf() situation, will produce a C4018 (signed/unsigned mismatch) warning at warning level 3 and above, beginning with C version 6.0.

Sample Code

   /* Compile options needed: none
   */

   #include <stdio.h>

   void main( void)
   {
      char goo = 127;
      char myvar = 128;

      printf( "goo = %X\n",   goo); /* yields "7F" */
      printf( "myvar = %X\n",   myvar); /* yields "FF80" or "FFFFFF80"*/

      /* Correct way to represent the char as two hex digits "80". */
      printf( "myvar = %X\n",   (unsigned char)myvar);
   }