Why MSVC generates warning C4127 when constant is used in “while” - C

Question

For code,

while(1)
{
   /* ..... */
}

MSVC generates the following warning.

warning C4127: conditional expression is const         


        

Answer 1

for(;;) and while (true) are different in that the former is a special case defined to be an infinite loop, while the latter is sort of an abuse saying "true, always."

The warning comes up because infinite loops when you don't want them are pretty bad, so it's warning you that you might have one at the first sign. But by using for(;;), you've pretty much explicitly said "loop this forever", and there's nothing to warn about.

I don't think GCC has an equivalent warning.

Answer 2

Constant conditionals are often enough simply bugs. Consider this:

unsigned k; 
...
while (k>=0) 
{
 ...
}

The condition k>=0 would make sense if k was a signed int, but not for unsigned. A careless developer forgets that k was declared unsigned and he/she would use it as if it was usable as a negative number. The compiler tries to be helpful and warn you about this and while(1) falls for the compiler into the same problem class. for(;;) is preferable because it unambiguously means `loop forever