Q67785: Bit Operations on Char May Produce Unexpected Results

Article: Q67785
Product(s): See article
Version(s): 6.00 6.00a | 6.00 6.00a
Operating System(s): MS-DOS | OS/2
Keyword(s): ENDUSER | buglist6.00 | mspl13_c
Last Modified: 6-FEB-1991

Bit manipulations on type char may produce different results depending
on the compiler switches. The /J and /qc switches have different
effects on the code below when using the Microsoft C versions 6.00 and
6.00a compiler.

Sample Code
-----------

#include <stdio.h>

char unsigned ary[4] = {0xF6, 0xF7, 0xF8, 0xF9};

void main(void)
{
   unsigned long result1, result2;
   char *p;
   p = ary;
   result1 = *p | ( *(p+2) << 8);
   result1 += ( *(p+1) | ( *(p+3) << 8)) * 0x10000;
   result2 = (unsigned long)(( *(p+2) << 8) | *p) |
             (unsigned long)(( *(p+3) << 8) | *(p+1)) << 16;
   printf("result1 = %lX\n", result1);
   printf("result2 = %lX\n", result2);
}

When compiled with

   cl /qc /J foo.c

the results are:

   result1 = F9F7F8F6
   result2 = F9F7F8F6

When compiled with

   cl /J foo.c

the results are:

   result1 = F9F6F8F6
   result2 = FFFFF8F6

When compiled with

   cl /qc foo.c
or
   cl foo.c

the results are:

   result1 = FFF6FFF6
   result2 = FFFFFFF6

The differences occur because of a problem in the Quick Compiler. When
an integral promotion is required, the Quick Compiler incorrectly
converts an unsigned char to an unsigned int. According to ANSI, if
all the values of a char can be represented in an int, it is converted
to an int; otherwise, it is converted to an unsigned int (see section
3.2.1.1). This is also documented in the "Advanced Programming
Techniques" manual on page 422.

There are a number of ways to work around this problem depending on
the desired results. If the intent was to generate results equal to
F9F7F8F6 (as the Quick Compiler with /J did), declare "p" as a pointer
to an unsigned char and modify the equations for "result1" and
"result2" to use unsigned int casts where appropriate. For example:

   result1 = *p | ((unsigned int)*(p+2) << 8);
   result1 += ( *(p+1) | ( *(p+3) << 8)) * 0x10000;

-or-

   result2 = ((unsigned long)(((unsigned int)*(p+2) << 8) | *p)|
             (unsigned long)(( *(p+3) << 8) | *(p+1)) << 16;

If the results from the full optimizing compiler with /J are desired,
again declare "p" as a pointer to unsigned char and replace the above
mentioned casts with signed int. Finally, if the results without /J
are desired, declare "p" as a pointer to a signed char.

Microsoft has confirmed this to be a problem in the Microsoft C
Compiler versions 6.00 and 6.00a. We are researching this problem and
will post new information here as it becomes available.

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