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

You can't. The preprocessor, as this names indicates, processes the source file before the compiler. It has therefore no knowledge of the values of your template arguments.

Answer 2

If you need to generate different code paths with template parameter, you can just simply use if or other C++ statement:

template <bool var>
void f()
{
    if (var == true) ｛
        // ...
    ｝
｝

Compiler can optimize it and generate code that doesn't contain such branches.

A little drawback is that some compiler (e.g. Msvc) will generate warnings for conditions which is always constant.

Answer 3

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

template <bool var>
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.

Answer 4

You can't do that with the preprocessor. All you can do is delegate the code to a separate template, something like this:

template <bool var>
void only_if_true()
{}

template <>
void only_if_true<true>()
{
  your_special_code_here();
}


template <bool var>
void f()
{
  some_code_always_used();
  only_if_true<var>();
  some_code_always_used();
}

Of course, if you need information shared between f() and only_if_true() (which is likely), you have to pass it as parameters. Or make only_if_true a class and store the shared data in it.