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

The fact that the result would be the same in most implementations in this case is incidental; the order of evaluation is still undefined. Consider f(i = -1, i = -2): here, order matters. The only reason it doesn't matter in your example is the accident that both values are -1.

Given that the expression is specified as one with an undefined behaviour, a maliciously compliant compiler might display an inappropriate image when you evaluate f(i = -1, i = -1) and abort the execution - and still be considered completely correct. Luckily, no compilers I am aware of do so.