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

The compiler will optimize n + 1 into nothingness.

Do you mean n = n + 1?

If so, they will compile to identical assembly. (Assuming that optimizations are on and that they're statements, not expressions)