At what point in the loop does integer overflow become undefined behavior?

Question

This is an example to illustrate my question which involves some much more complicated code that I can\'t post here.

#include 
int main()
{
           


        

Answer 1

Technically, under the C++ standard, if a program contains undefined behavior, the behavior of the entire program, even at compile time (before the program is even executed), is undefined.

In practice, because the compiler may assume (as part of an optimization) that the overflow will not occur, at least the behavior of the program on the third iteration of the loop (assuming a 32-bit machine) will be undefined, though it is likely that you will get correct results before the third iteration. However, since the behavior of the entire program is technically undefined, there's nothing stopping the program from generating completely incorrect output (including no output), crashing at runtime at any point during execution, or even failing to compile altogether (as undefined behavior extends to compile time).

Undefined behavior provides the compiler with more room to optimize because they eliminate certain assumptions about what the code must do. In doing so, programs that rely on assumptions involving undefined behavior are not guaranteed to work as expected. As such, you should not rely on any particular behavior that is considered undefined per the C++ standard.