Preprocessor and template arguments or conditional compilation of piece of code

Question

How I can compile template function with pre-processor condition? Like that (but it is not working):

template 
void f()
{
    #if (var == tru         


        

Answer 1

With C++17's introduction of if constexpr you can discard branches inside a template, much like conditional compilation allows.

template 
void f()
{
    if constexpr (var == true) {
    // ...
    }
}

The code inside the branch has to be syntactically correct, but doesn't have to be well-formed when var is false, because it will be discarded entirely.