Q60734: C4127 Is Generated When Subexpression Evaluates to a Constant

Article: Q60734
Product(s): See article
Version(s): 6.00   | 6.00
Operating System(s): MS-DOS | OS/2
Keyword(s): ENDUSER | | mspl13_c
Last Modified: 21-APR-1990

Microsoft C Version 6.00 has a new compiler warning "C4127:
conditional expression is constant." This warning is designed to
inform you that the controlling expression of an if statement or while
loop evaluates to a constant, so the body of the loop is ALWAYS
executed or NEVER executed.

The warning may appear in certain expressions that don't seem to be a
constant, but this is because the compiler will generate this warning
if ANY subexpression in a larger conditional expression evaluates to a
constant. The warning is strictly informational and does not
necessarily indicate any problems in the code.

In the sample code below, warning C4127 is generated if the code is
compiled at warning level three or four (/W3 or /W4). Looking at the
following entire expression

     ( hours >= 0 && hours <= 24 )

it definitely appears that it is NOT a constant because hours could be
EITHER in the range 0 (zero) to 24, or out of that range. This
expression generates warning C4127, however, because the left
subexpression

      hours >= 0

ALWAYS evaluates to true because hours is unsigned and an unsigned int
is ALWAYS greater than or equal to zero. The compiler generates the
warning to inform you of this situation.

Note that this warning is generated only by the full optimizing
compiler because the quick compiler (/qc) does not check for this
situation.

Sample Code:

#include <stdio.h>

void main(void)
{
    unsigned hours;

    scanf ( "%ud", &hours );

    if ( hours >= 0 && hours <= 24 )
        printf("Hours OK\n");
    else
        printf("Hours BAD\n");
}

Making a simple change, such as replacing the ">=" with a ">",
eliminates the warning because the left expression could now evaluate
to either true or false (for example, false if hours = 0; true
otherwise).

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