Does the compiler remove if statements where the condition is always false? [duplicate]

Question

I know some metaprogramming techniques in C++ to compute constants at compile-time. Most of the time branching in metafunctions is done through the ternary operators which can be evaluated at compile-time on the contrary of standard if/else.

But regarding this kind of function:

template <unsigned int N>
void f()
{
    if (N == 0) {
        // Some computations here
    } else if (N <= 42) {
        // Some computations here
    } else {
        // Some computations here
    }
}

What will the compiler do (assuming -O3) ? The compiler knows that f<0>() will always branch on the first case, f<32>() will always branch on the second case, and f<64>() will always branch on the third case.

Will the compiler remove branches that will always be false ? Will it directly branch to the only valid case ?

Answer 1

The optimizer will remove the branch and the code in the unused branches, but beware: the compiler needs to process the function before the optimizer even gets a chance to look at the code, which means that all branches must be valid (compilable) for all values of N.

For example, if the second branch contained:

} else if (N <= 42) {
   char data[50 - N];
// other code

The compiler will fail to instantiate the template for N >= 50 even though the branch will be removed by the optimizer.