Yes, it'll exploit the fact that signed overflow is undefined. int incr(int i) { int old = i; i += 1; if (i < old) return 1; return

Yes, it'll exploit the fact that signed overflow is undefined. int incr(int i) { int old = i; i += 1; if (i < old) return 1; return 0; } GCC optimizes this to 'return 0;'. EDIT: Updated to a better example.