Variadic Template Functor Call

Question

So I've been trying to use variadic templates to compose objects out of more convenient subtypes, but I'm having trouble getting it to do exactly what I want.

template<class ...Functor>
struct SeqMethod:public Functor...{
  template<class F>
  void call(F& a){
    F::operator()();
  }
  template<class F,class ... funcs>
  void call(){
    F::operator()();

    call<funcs...>();
  }
  public:
  void operator()(){
    call<Functor...>();
  }
};

This isn't valid syntax, so there's that.

Ideally I'd like to be able to use something like this

class A{
public:
  void operator()(){
    std::cout<<"A";
  }
};
class B{
public:
  void operator()(){
    std::cout<<"B";
  }
};

class C:public SeqMethod<A,B>{};

Which in this case should output "AB", and in general be suitable for composing behaviors together.

skypjack · Answer 1 · 2016-12-24T00:02:54.697

6

You don't actually need any call member function in your case.
Instead you can do this in C++11/C++14:

template<class ...Functor>
struct SeqMethod:public Functor...{
  public:
  void operator()(){
      int _[] = { (Functor::operator()(), 0)... };
      return void(_);
  }
};

It follows a minimal, working example:

#include<iostream>

template<class ...Functor>
struct SeqMethod:public Functor...{
  public:
  void operator()(){
    int _[] = { (Functor::operator()(), 0)... };
    return void(_);
  }
};

class A{
public:
  void operator()(){
    std::cout<<"A";
  }
};
class B{
public:
  void operator()(){
    std::cout<<"B";
  }
};

class C:public SeqMethod<A,B>{};

int main() {
  C c;
  c();
}

edited Dec 24 '16 at 00:02

answered Dec 23 '16 at 23:50

skypjack

49,335
19
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187

1

C++17 changed things a bit. This is the C++14 way to unroll a parameters pack. SO as well as the web are full of in-depth explanations for it, more detailed than what I could in a comment. ;-) – skypjack May 23 '19 at 13:34
thanks for quick reply. I know about the fold expression in C++17. But currently I'm using C++11, and all i could find is implementation like question. Can you send some keywords i should look for and is it possible to do this without declaring an array ? Thanks again. – s0nskar May 23 '19 at 13:52
1

You should search for unpacking parameter pack in C++11 or something like this. In general, an array like thing is required btw. – skypjack May 23 '19 at 13:53

score 3 · Accepted Answer · answered Dec 23 '16 at 23:50

3

The easiest way to do this is with C++17's fold expressions:

template<class ...Functor>
struct SeqMethod:public Functor...{

 public:
    void operator()(){
        (Functor::operator()(),...);
    }
};

class A{
public:
    void operator()(){
        std::cout<<"A";
    }
};
class B{
public:
    void operator()(){
        std::cout<<"B";
    }
};

class C:public SeqMethod<A,B>{};

int main()
{
    C c;
    c();
    return 0;
}

Output (tested with gcc 6.2):

AB

answered Dec 23 '16 at 23:50

Sam Varshavchik

114,536
5
94
148

This is great, I really appreciate this option, but I'm working on some platforms whose c++ compilers are a bit outdated, so since the other option(the initializer list hack) is a bit easier to come by as far as support goes. – violet_white Dec 24 '16 at 00:01

Variadic Template Functor Call

2 Answers2