What's the point of issuing a compiler warning for "while(true)" and not issuing one for "for(;;)"?

Question

When I compile C++ code with Visual C++ 9 with "warning level 4" the following:

while( true ) {
   //loop body with break on certain condition
}

and the following:

for( ; true; ) {
   //same loop body
}

both trigger C4127: conditional expression is constant warning but the following:

for( ; ; ) {
   //same loop body
}

compiles without warning.

Why this difference, especially between the second and the third variant?

Answer 1

The reason for warning the user of constant conditional expressions is to help avoid bugs where the expression ends up being constant (for example, due to a typo). In the last case, there is no expression, so there is no risk of it accidentally being constant.

Answer 2

The reason is simple, though stupid.

It is important to diagnose infinite loop, but such might not be evident:

while(i >= 0) { --i; } // infinite if i unsigned

while(SOME_MACRO(i)) // err, depends on the expansion of macros

It is a great feature of a compiler to produce a warning for a tautological test, that is a test that turns out to be either always true or always false, because it's not obvious when it comes from a macro expansion or within a dependent context.

It just seems that VC++ pushed a bit too far here, and instead of considering tautological conditions warns for all true or false conditions it can find, even when they are already clearly stated in the code.