[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 in the 'test' position)
Saves usually a register and replaces a check against an 'arbitrary'
number with a check for zero (which is faster/smaller on some
architectures).
Of course, this won't work when you can't invert the iteration order.
Andre'
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