How to pass around parameter packs in C++?

Question

Consider the following example:

template <class T> class method_traits;
template <class T, class Ret, class... Arg> class method_traits<Ret(T::*)(Arg...)> {
public:
    using type = Arg; // this does not work
};

template <class T> using argument_types = typename method_traits<T>::type;

template <class T> class Node {
    T t;
public:
    Node(Input<argument_types<decltype(&T::process)>>... inputs) { // how do I make this work?
        ...
    }
};

The arguments of the constructor of Node<T> depend on the arguments of the method T::process. So if a type T has a method process of the signature float process(float a, int b) the signature of the constructor of Node<T> should look like this: Node(Input<float> a, Input<int> b).

How do I extract the parameter pack from T::process to use it on the constructor of Node?

Gemmagemmate

You need to make Node a variadic template too. – Burrill

@n.m. but Node only depends on a single type T and not on multiple types – Gemmagemmate

@n.m. Surely it suffices to make Node's constructor a variadic template, rather than the whole class? – Collegiate

@ArthurTacca it could be, depending on what Node is storing. – Burrill

In my case Node stores the Inputs in a tuple so I guess making the constructor a variadic template would work. – Gemmagemmate

You cannot say "my member function will have the same signature as T::somefunction. It's a deficiency in C++ syntax. To overcome it, you need to introduce an artificial parameter pack that doesn't express any real dependency. – Burrill

What's Input? Is that important to the question? – Uncourtly

Note most of the answers will fail if the member function has cv-qualifiers, has ref-qualifiers, or is C-style variadic, (or in C++17, is noexcept). See the ugly sample definition of std::is_function at en.cppreference.com/w/cpp/types/is_function for a hint on how it would need to be done properly. – Marmara

@Uncourtly It is not important to the question. It is just there to show that the arguments are not exactly the same but depend on each other. – Gemmagemmate

Answer 1

Obviously you can't save a list of types in this way

    using type = Arg;

where Arg is a variadic list of types.

But you can save they in a type container and std::tuple can do this works too. So I suggest to modify the method_traits specialization as follows

template <typename T>
struct method_traits;

template <typename T, typename Ret, typename... Args>
struct method_traits<Ret(T::*)(Args...)>
 { using tTypes = std::tuple<Args...>; };

and rewrite argument_types to intercept the std::tuple

template <typename T>
using tTypes = typename method_traits<T>::tTypes;

Now you can use the default template value and partial specialization trick defining node

template <typename T, typename TArgs = tTypes<decltype(&T::process)>>
struct Node;

In this way, instantiating a Node<T> object, you effectively get a Node<T, tTypes<decltype(&T::process)> that is a Node<T, std::tuple<Args...>> with the wanted Args....

So you can simply define the following partial specialization of Node as follows

template <typename T, typename ... Args>
struct Node<T, std::tuple<Args...>>
 {
   T t;

   Node (Input<Args> ... inputs)
    { /* do something */ }
 };

The following is a full working example

#include <tuple>
#include <type_traits>

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

template <typename T>
using Input = typename tWrapper<T>::type;

template <typename T>
struct method_traits;

template <typename T, typename Ret, typename... Args>
struct method_traits<Ret(T::*)(Args...)>
 { using tTypes = std::tuple<Args...>; };

template <typename T>
using tTypes = typename method_traits<T>::tTypes;

template <typename T, typename TArgs = tTypes<decltype(&T::process)>>
struct Node;

template <typename T, typename ... Args>
struct Node<T, std::tuple<Args...>>
 {
   T t;

   Node (Input<Args> ... inputs)
    { /* do something */ }
 };

struct foo
 {
   float process (float a, int b)
    { return a+b; }
 };

int main ()
 {
   Node<foo> nf(1.0f, 2);
 }

-- EDIT --

As pointed by Julius (and the OP themselves) this solution require an additional template type with a default template value.

In this simplified case isn't a problem but I can imagine circumstances where this additional template argument can't be added (by example: if Node receive a variadic list of template arguments).

In those cases, Julius propose a way that complicate a little the solution but permit to avoid the additional template parameter for Node: to add a template base class, that receive the TArgs arguments, and to works with constructor inheritance.

That is: defining a NodeBase as follows

template <typename, typename>
struct NodeBase;

template <typename T, typename ... Args>
struct NodeBase<T, std::tuple<Args...>>
 {
   T t;

   NodeBase (Input<Args> ...)
    { /* do something */ }
 };

there is no need for an additional template parameter, for Node, that can simply written as

template <typename T>
struct Node
   : public NodeBase<T, tTypes<decltype(&T::process)>>
 { using NodeBase<T, tTypes<decltype(&T::process)>>::NodeBase; };

Julius, following this idea, prepared a solution that (IMHO) is even better and interesting.

Nephralgia

nice solution, even though it requires to add a second template parameter to Node – Gemmagemmate

@Gemmagemmate - yes; but the second parameter is a default one; so, as you can see in main(), you can instantiate a Node simply as Node<foo>. – Nephralgia

Very nice solution. If that default template parameter is not acceptable, one could derive from a base class with constructor inheritance. – Lefty

@Lefty - thanks; I prefer the default template type way, but I can imagine circumstances where it's an impracticable solution. In that circumstances, your idea can be useful. – Nephralgia

Answer 2

Using perfect forwarding (live):

template<typename... Args>
Node(Args&&... args) {
    process(std::forward<Args>(args)...);
}

Gruesome

Honestly this is the answer, all the answers above are needlessly convoluted. – Beloved

Answer 3

Here is one example in C++11 (thanks to max66's comment) based on max66's great answer. The differences here:

• no additional template argument for Node (instead, using a base class and constructor inheritance)
• the desired arguments are obtained with a slightly different style than shown in the question
  • adding overloads for qualified member functions is simple
  • references are not a problem (as far as I can tell; see example below printing 42)

http://coliru.stacked-crooked.com/a/53c23e1e9774490c

#include <iostream>

template<class... Ts> struct Types {};

template<class R, class C, class... Args>
constexpr Types<Args...> get_argtypes_of(R (C::*)(Args...)) {
    return Types<Args...>{};
}

template<class R, class C, class... Args>
constexpr Types<Args...> get_argtypes_of(R (C::*)(Args...) const) {
    return Types<Args...>{};
}

template<class T, class ConstructorArgs>
struct NodeImpl;

template<class T, class Arg0, class... Args>
struct NodeImpl<T, Types<Arg0, Args...>> {
  NodeImpl(Arg0 v0, Args...) {
    v0 = typename std::decay<Arg0>::type(42);
    (void)v0;
  }
};

template<class T>
struct Node
  : NodeImpl<T, decltype(get_argtypes_of(&T::process))>
{
  using ConstructorArgs = decltype(get_argtypes_of(&T::process));
  using NodeImpl<T, ConstructorArgs>::NodeImpl;
};

struct Foo {
    void process(int, char, unsigned) const {}
};

struct Bar {
    void process(double&) {}
};

int main() {
    Node<Foo> foo_node{4, 'c', 8u};

    double reftest = 2.0;
    Node<Bar> bar_node{reftest};
    std::cout << reftest << std::endl;
}

Lefty

Also your's is a nice solution; if you want it to works with C++11, it's very simple: avoid to return auto in get_argtypes_of() and explicit that return a Types<Args...>. If you're worried for the repetition of Types<Args...>, you can simply return {};. (continue) – Nephralgia

And, considering that you use get_argtypes_of() only in a couple of decltypes() (because you're only interested in returned type, non in returned value) you can also declare it and not implement it (as std::declval()); I mean that it's enough template<class R, class C, class... Args> constexpr auto get_argtypes_of(R (C::*)(Args...));, without the implementation of the function. – Nephralgia

Answer 4

Here is a solution using C++14. (Note: I have only tested it in clang, though):

#include <string>
#include <utility>

struct Foo {
    void process(int, std::string);
};

template <typename T>
struct Input { };

template <std::size_t N, typename T, typename ...Types>
struct Extract_type {
    using type = typename Extract_type<N - 1, Types...>::type;
};

template <typename T, typename ...Types>
struct Extract_type<0, T, Types...> {
    using type = T;
};

template <typename T, std::size_t N, typename R, typename ...Args>
typename Extract_type<N, Args...>::type extract(R (T::*)(Args...));

template <typename T, typename R, typename ...Args>
std::integral_constant<std::size_t, sizeof...(Args)> num_args(R (T::*)(Args...));

template <typename T>
struct Node {
    template <typename ...Args>
    Node(Input<Args>&&... args) 
    : Node(std::make_index_sequence<decltype(num_args<T>(&T::process))::value>{ }, std::forward<Input<Args>>(args)...)
    {}

    template <std::size_t ...Indices>
    Node(std::index_sequence<Indices...>, Input<decltype(extract<T, Indices>(&T::process))>...) {}
};


int main() {
    Node<Foo> b{ Input<int>{ }, Input<std::string>{ } };
}

http://coliru.stacked-crooked.com/a/da7670f80a229931

Chace

Answer 5

How about using private inheritance and CRTP?

#include <tuple>
#include <iostream>

template <typename Method> struct method_traits;

template <typename T, typename Ret, typename... Args>
struct method_traits<Ret(T::*)(Args...)> {
public:
    using parameter_pack = std::tuple<Args...>;
};

template <typename Derived, typename Tuple> struct Base;

template <typename Derived, typename... Ts>
struct Base<Derived, std::tuple<Ts...>> {
    void execute_constructor(Ts&&... ts) { 
        Derived* d = static_cast<Derived*>(this);
        d->t.process(std::forward<Ts>(ts)...);
        d->num = sizeof...(Ts);
    }
    virtual ~Base() = default;
};

template <typename T, typename... Rest>
class Node : Base<Node<T, Rest...>, typename method_traits<decltype(&T::process)>::parameter_pack> {
    T t;
    int num;
public:
    using Base = Base<Node<T, Rest...>, typename method_traits<decltype(&T::process)>::parameter_pack>;
    friend Base;  // So that Base can do whatever it needs to Node<T, Rest...>'s data members.
    template <typename... Ts>
    Node (Ts&&... ts) {
        Base::execute_constructor(std::forward<Ts>(ts)...);
        std::cout << "num = " << num << '\n';
    }
};

struct foo {
    void process(int a, char c, bool b) {
        std::cout << "foo(" << a << ", " << c << ", " << std::boolalpha << b << ") carried out.\n";
    }   
};

int main() {
    Node<foo> n(5, 'a', true);
    std::cin.get();
}

Output:

foo(5, a, true) carried out.
num = 3

Kellene