Why does n++ execute faster than n=n+1?

Question

In C language, Why does n++ execute faster than n=n+1?

(int n=...;  n++;)
(int n=...;  n=n+1;)

Our instructor asked

Answer 1

In C language the side-effect of n++ expressions is by definition equivalent to the side effect of n = n + 1 expression. Since your code relies on the side-effects only, it is immediately obvious that the correct answer is that these expression always have exactly equivalent performance. (Regardless of any optimization settings in the compiler, BTW, since the issue has absolutely nothing to do with any optimizations.)

Any practical divergence in performance of these expressions is only possible if the compiler is intentionally (and maliciously!) trying to introduce that divergence. But in this case it can go either way, of course, i.e. whichever way the compiler's author wanted to skew it.