How to make template parameter

问题:

How do I create a metafunction that takes any kind of function pointer? In the code below, how do I get rid of "decltype(&f)" ?

template <class FuncType, FuncType functionPointer>
void runFunc()
{
    functionPointer();
}

runFunc<decltype(&f),f>();

I don't want to have to specify the type of f seperately; the information is already there in f. I'd prefer not to resort to defines to solve this. This is basically the templated function type idiom applied to meta-programming; I don't want to know the type of f, but whatever I get in apparently allows me to call operator() on it.

Stuff I've tried:

Different order for the template parameters; since later parameters seem to be guessable when you have a function; not possible because then you would need to forward declare FuncType, in order to have it as the type for functionPointer

Switching it around so that you specify returntype and parameters and give a function pointer of that type; cannot instantiate a template with variable template arguments in the middle; that looks like below:

template <class ReturnType, class ... ArgTypes, ReturnType (*functionPointer)(ArgTypes...)>
void runFunc()
{
    functionPointer();
}

runFunc<void, int, f>(); // error; invalid template argument for 'ArgTypes', type expected

There's a bit more context here on github: https://github.com/TamaHobbit/FuncTest/blob/master/FuncTest/FuncTest.cpp

回答1:

Right now there is unfortunately no good way of doing this.

The standard committee has, however, accepted a proposal that makes this valid code:

template <auto functionPointer>
void runFunc() {
  functionPointer();
}

Compiler support should be coming soon.

回答2:

You can use this:

template <typename FuncType>
void runFunc(FuncType functionPointer )
{
    functionPointer();
}

runFunc(f);