Signed overflow in C++ and undefined behaviour (UB)

Question

Compilers do assume that a valid C++ program does not contain UB. Consider for example:

if (x == nullptr) {
    *x = 3;
} else {
    *x = 5;
}

If x == nullptr then dereferencing it and assigning a value is UB. Hence the only way this could end in a valid program is when x == nullptr will never yield true and the compiler can assume under the as if rule, the above is equivalent to:

*x = 5;

Now in your code

int result = 0;
int factor = 1;
for (...) {      // Loop until factor overflows but not more
   result = ...
   factor *= 10;
}
return result;

The last multiplication of factor cannot happen in a valid program (signed overflow is undefined). Hence also the assignment to result cannot happen. As there is no way to branch before the last iteration also the previous iteration cannot happen. Eventually, the part of code that is correct (i.e., no undefined behaviour ever happens) is:

// nothing :(

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