3 ms·

The inequality argument would seem to apply only to for loops with constant iteration counts, yeah? The compiler can't know that "i < n; i++" won't overflow if

• dooglius 4y ago
  If "i" started out higher than "n", the loop would exit immediately and there would be no increment. More generally, if "n" is an int, then even if n is modified each iteration, the "i < n; i++" still can't overflow because it will always exit at INT_MAX if not before. If "n" is something else like a long then that's likely a mistake that falls under "the compiler should warn if you do this". If there were good reason for wanting to assume this, I would expect someone to have presented it; responses I hear just take the form of a single (usually contrived) example where more performant code is generated, then insisting that all other factors (e.g. being predictable/unsurprising, acting like the underlying architecture, mitigating bugs) are irrelevant. Just as GP did.
  • kalkin 4y ago
    Ah good point re the inequality, over thought it. If there weren't good reasons for making the assumptions that modern compilers do, I'd have expected users of C and C++ to have converged on a more conservative compiler, at least in some significant number.