I know the (non-bitwise) boolean operators have shortcutting behavior, but if evaluating the second clause has no side effects, couldn't the compiler just cheat

I know the (non-bitwise) boolean operators have shortcutting behavior, but if evaluating the second clause has no side effects, couldn't the compiler just cheat by evaluating both clauses unconditionally and then combining the two boolean results using the equivalent bitwise operation, thus avoiding a branch?