Expand parameter pack for a function template

Question

I have a function template as below. The template argument needs to be explicitly given.

template<typename T>
void Func() {...};

I need to call this function for each type in a parameter pack:

template<typename... Inputs>
struct SetStruct{
  void Set() {
    // Call Func() here
  }
};

Is there an easy way to expand the parameter pack? I tried:

Func<Inputs>()...;

and

Func<Inputs>...();

But none of them works.

I can only use C++11 :(

OP doesn't specify C++11, other question was a C++11 question, not a duplicate. — Yakk - Adam Nevraumont, Nov 09 '18 at 17:01
@Yakk-AdamNevraumont Variadic template param packs imply c++11 as a minimum. — πάντα ῥεῖ, Nov 09 '18 at 17:02
@πάνταῥεῖ Now it is a duplicate, it wasn't before. Tags matter. — Yakk - Adam Nevraumont, Nov 09 '18 at 17:27

score 5 · Accepted Answer · edited Jun 20 '20 at 09:12

Is there an easy way to expand the parameter pack? I tried:

Func<Inputs>()...;

If you can use C++17, using the comma operator and template-folding

((void)Func<Inputs>(), ...);

In C++11/C++14, using again the comma operator but in the context of initialization of an unused C-style array, something as follows

template<typename... Inputs>
struct SetStruct{
  void Set() {
    using unused = int[];

    (void)unused { 0, ((void)Func<Inputs>(), 0)... };
  }
};

Observe that, in both cases, I've added a (void) before the call to Func<>().

In your case it's useless (because your Func<>() just return void) but it's a sort of security belt in case of a function that return an object of a class that redefine the comma operator.

Expand parameter pack for a function template

1 Answers1