iterating over variadic template's type parameters

Question

I have a function template like this:

template 
do_something()
{
  // i\'d like to do something to each A::var, where var has static storag         


        

Answer 1

What Xeo said. To create a context for pack expansion I used the argument list of a function that does nothing (dummy):

#include <iostream>
#include <initializer_list>

template<class...A>
void dummy(A&&...)
{
}

template <class ...A>
void do_something()
{
    dummy( (A::var = 1)... ); // set each var to 1

    // alternatively, we can use a lambda:

    [](...){ }((A::var = 1)...);

    // or std::initializer list, with guaranteed left-to-right
    // order of evaluation and associated side effects

    auto list = {(A::var = 1)...};
}

struct S1 { static int var; }; int S1::var = 0;
struct S2 { static int var; }; int S2::var = 0;
struct S3 { static int var; }; int S3::var = 0;

int main()
{
    do_something<S1,S2,S3>();
    std::cout << S1::var << S2::var << S3::var;
}

This program prints 111.

Answer 2

As an example, suppose you want to display each A::var. I see three ways to acomplish this as the code below illustrates.

Regarding option 2, notice that the order in which the elements are processed is not specified by the standard.

#include <iostream>
#include <initializer_list>

template <int i>
struct Int {
    static const int var = i;
};

template <typename T>
void do_something(std::initializer_list<T> list) {
    for (auto i : list)
        std::cout << i << std::endl;
}

template <class... A>
void expand(A&&...) {
}

template <class... A>
void do_something() {

    // 1st option:
    do_something({ A::var... });

    // 2nd option:
    expand((std::cout << A::var << std::endl)...);

    // 3rd option:
    {
        int x[] = { (std::cout << A::var << std::endl, 0)... };
        (void) x;
    }
}

int main() {
    do_something<Int<1>, Int<2>, Int<3>>();
}