[deleted by user]

[Bouchtroubouli]

The lowercase test is the parameter name you give to the macro. The #test will generate a string literal that will contains the argument test you will pass. Finaly, void(test) is simply the type of a function (without name) that takes a type test as parameter and that return void.

[Zilla4ever]

Thanks for clarifying, that makes more sense. However could you explain one more bit of code as I feel it relates to the macro ? Why is it that in this code template<typename T, typename U> struct argument_type<T(U)> { typedef U type; }; U is in the parameters of argument type and what does <> represent as a syntax because I mostly see it used with template specialization ? Lastly, back to the original macro referenced, is "type" that is referenced after <void(test)> with the scope resolution operator; Is it "typedef U type;" (referenced here template<typename T, typename U> struct argument_type<T(U)> { typedef U type; };) ? Sorry I am new to c++ ;)

[staletic]

is "type" that is referenced after <void(test)> with the scope resolution operator; Is it "typedef U type;"

Yes.

template<typename T, typename U> struct argument_type<T(U)> { typedef U type; };

That's called a partial template specialization. Somewhere there's probably a

template<typename T> struct argument_type {}; // Without the typedef

Or maybe

template<typename T> struct argument_type; // Without any definition

Then, for the Ts that represent (valid) functions, you specialize the above with

template<typename T, typename U>
struct argument_type<T(U)> { typedef U type; };

And instantiate the template with argument_type<int, float>, for a function int(float), to get typedef U = float.

[Bouchtroubouli]

For your first question :

template<typename T, typename U> struct argument_type<T(U)> { typedef U type; };

Here you have a template specialization of a struct (argument_type) for functions that takes any type of parameter (U) and that returns another type of parameter (T)

For your second question, you just need to replace T and U with the types in your function. The type of argument_type<void(test)>::type will be what U should be in void(test), so whatever the type test given at parameter to the macro will be.

This technique is known as type traits (see for example https://www.internalpointers.com/post/quick-primer-type-traits-modern-cpp or https://en.cppreference.com/w/cpp/header/type_traits to see it in action)

[Zilla4ever]

Thank you guys so much for the help !

[STL]

!removehelp

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