Why is f(i = -1, i = -1) undefined behavior?

Question

I was reading about order of evaluation violations, and they give an example that puzzles me.

1) If a side effect on a scalar object is un-sequenced r

Answer 1

Actually, there's a reason not to depend on the fact that compiler will check that i is assigned with the same value twice, so that it's possible to replace it with single assignment. What if we have some expressions?

void g(int a, int b, int c, int n) {
    int i;
    // hey, compiler has to prove Fermat's theorem now!
    f(i = 1, i = (ipow(a, n) + ipow(b, n) == ipow(c, n)));
}