if (num < a) return num * num;
/*else*/ return num;
If the code returns `num * a` then the value of `a` is now necessary, and must be computed before the function returns.
For signed integer addition the compiler is allowed to assume that `(num < (num + 1))` is true, so the comparison can be removed entirely.
That's not directly what the compiler assumes. The direct problem is in 'a + 1' having undefined behaviour, and that transitively allows the assumption on the comparison that you mentioned.
This was an example where 'a + 1' doesn't compile to an add instruction.