For signed int, the wrap around behavior is undefined. If we allow !=, programmers may get unexpected tripcount. The problem is whether the compiler can generate code to compute a trip count for the loop.
For a simple loop, like:
'for( i = 0; i < n; ++i )'
the compiler can determine that there are 'n' iterations, if n>=0, and zero iterations if n < 0.
For a loop like:
'for( i = 0; i != n; ++i ) '
again, a compiler should be able to determine that there are 'n' iterations, if n >= 0; if n < 0, we don't know how many iterations it has.
For a loop like:
for( i = 0; i < n; i += 2 )
the compiler can generate code to compute the trip count (loop iteration count) as floor((n+1)/2) if n >= 0, and 0 if n < 0.
For a loop like:
for( i = 0; i != n; i += 2 )
the compiler can't determine whether 'i' will ever hit 'n'. What if 'n' is an odd number?
For a loop like:
for( i = 0; i < n; i += k )
the compiler can generate code to compute the trip count as floor((n+k-1)/k) if n >= 0, and 0 if n < 0, because the compiler knows that the loop must count up; in this case, if k < 0, it's not a legal OpenMP program.
For a loop like:
for( i = 0; i != n; i += k )
the compiler doesn't even know if i is counting up or down. It doesn't know if 'i' will ever hit 'n'. It may be an infinite loop.
