Q51472: Shift Left Assignment Operator "<<=" Does Not Work Correctly

Article: Q51472
Product(s): See article
Version(s): 2.00 2.01
Operating System(s): MS-DOS
Keyword(s): ENDUSER | S_QUICKC S_QUICKASM buglist2.00 buglist2.01 | mspl13_c
Last Modified: 17-JAN-1990

The shift left assignment operator, "<<=", does not operate correctly
on variables that are not in the default data segment in the small and
medium memory models.

The following code reproduces the problem:

#include<stdio.h>

/* must be compiled in small or medium memory model */

/* variables must be declared global so that "far"  */
/* modifier will not be ignored                     */

int a = 1;           /*  int "a" will be in default data segment */
int far b = 1;       /*  int "b" declared "far" in far segment   */

void main( void )
{
   a <<= 4;          /*  left shift 4 of 1 should be 16          */
   b <<= 4;          /*  this operation causes bad value         */
   /*  output results  */
   printf( "A: %d (should be 16)\nB: %d (should be 16)\n", a, b );
}

Output:

A: 16 (should be 16)
B: 1 (should be 16)

The reason that the bad value is returned is that the left shift
operation always assumes a near pointer in the small and medium memory
models, i.e., only an OFFSET value. When performed on a far value, the
address of that value is truncated to be only an OFFSET instead of
SEGMENT:OFFSET, and the segment becomes the default data segment. The
resulting pointer points to an unpredictable point in that data
segment, and the operation is carried out there instead of on the
intended value. In the large and compact memory models, the shift
operation performs as desired.

To work around the problem, the program should be recompiled in the
large or compact memory model, or the left shift operation should be
rewritten as follows:

   b = b << 4;

Microsoft has confirmed this to be a problem with the QuickC compiler
Versions 2.00 and 2.01. We are researching this problem and will post
new information here as it becomes available.

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