[EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> If your f() returns a constant, your best bet is to do:
>
> for (int i =0, j = f(n); i < j; ++i) {
> }
Or if it doen't matter in what order the iteration is performed,
even
for (int i = f(n); i--; )
might do. (Note the i-- is post-decrement and i
If you're doing more powerful optimizations, such as IPA based, its
likely you will be doing loop unrolling as well. And after loop
unrolling, they should generate code identical code.
> Saves usually a register and replaces a check against an 'arbitrary'
> number with a check for zero (which is f
