CODEWITHSUNDEEP
c++templatesparametersfunction-pointersusing
Jonathan Mee
Jonathan Mee

Reputation: 38919

Preserving Function Pointer Template Arguments

Typically I'll preserve function arguments something like this:

template<typename T>
struct foo {
    using type = T;
};

I know it's not going to be using but in what way can I preserve a function passed as a template parameter?

template<void (*T)(const int)>
struct bar {
    static T& type = T;
};

I know this doesn't work, I just want to know how I can go about the act of preserving this function pointer.

Upvotes: 0

Views: 51

Answers (2)

NathanOliver
NathanOliver

Reputation: 180710

You can use decltype to declare a member variable of the type of T that people can refer to like

template<void (*T)(const int)>
struct bar {
    static constexpr decltype(T) value = T;
};

You can take it a step farther and make the type of T a member as well like

template<void (*T)(const int)>
struct bar {
    using fptr = decltype(T)
    static constexpr fptr value = T;
};

Upvotes: 3

SergeyA
SergeyA

Reputation: 62583

When used like this:

template<void (*T)(const int)>
struct bar {
    static T& type = T;
};

T is a non-type template parameter, and there is no type to preserve (I would also suggest a different name for it, since T usually refers to type)

You can make a pointer itself available as a member of the struct, for example

static constexpr int (*func_ptr)(int) = T;

Upvotes: 2

Related Questions