Comment by caf

3 years ago

Yes - and the original post here is the same:

    if (x < 0)
        return 0;

The compiler now knows x's possible range is non-negative.

    int32_t i = x * 0x1ff / 0xffff;

A non-negative multiplied and divided by positive numbers means that i's possible range is also non-negative (this is where the undefinedness of integer overflow comes in - x * 0x1ff can't have a negative result without overflow occurring).

    if (i >= 0 && i < sizeof(tab)) {

The first conditional is trivially true now, because of our established bounds on i, so it can just be replaced with "true". This is what causes the code to behave contrary to the OP's expectations: with his execution environment in the overflow case we can end up with a negative value in i.

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