Question: Does the compiler still throw the warning about "x and y unused" when, instead of replacing z = x + 2*y + 1; by z = 5; // x + 2*y + 1; you do

Question: Does the compiler still throw the warning about "x and y unused" when, instead of replacing z = x + 2*y + 1; by z = 5; // x + 2*y + 1; you do something like z = (x + 2*y + 1) * 0 + 5;