converting parameter pack into a vector

Question

I am trying to understand variadic templates in C++ and I am lost in the following example: Imagine a function foo(T, T, T, T) which takes variadic number of arguments of the same type T and converts them into a vector. Any idea how to implement one?

It should work like this

foo<int>(1,2,3,4) returns std::vector<int> x{1,2,3,4}
foo<double>(0.1,0.2,0.3) returns std::vector<double> x{0.1,0.2,0.3}

Possible duplicate of https://stackoverflow.com/questions/13978916/inserting-a-variadic-argument-list-into-a-vector — , Nov 05 '17 at 20:27
`Is` is the parameter pack but you're trying to expand `i`. Is that the problem you're having or is it something else? Might want to try clarifying exactly what you're having a problem with. — Captain Obvlious, Nov 05 '17 at 20:38

max66 · Answer 1 · 2017-11-05T20:43:28.763

If the T values as known at compile-time, you can pass they as template parameters and write something like

template<typename T, T ... Is>
void foo() {
   std::vector<T> x { { Is.. } };

   for( auto xx:x )
      std::cout << xx << std::endl;
}

that is called

foo<int, 2, 3, 5, 7>();

Otherwise you have to pass they as arguments; something like

template <typename T, typename ... ARGS>
void foo (ARGS const & ... args) {    
   std::vector<T> x { { args... } };

   for( auto xx:x ) 
      std::cout << xx << std::endl;
}

that is called

foo<int>(2, 3, 5, 7);

or also (deducing the type T from the first argument)

template <typename T, typename ... ARGS>
void foo (T const & arg0, ARGS const & ... args) {    
   std::vector<T> x { { arg0, args... } };

   for( auto xx:x ) 
      std::cout << xx << std::endl;
}

that is called

foo(2, 3, 5, 7);

-- EDIT --

The OP write

It should work like this

foo<int>(1,2,3,4) returns std::vector<int> x{1,2,3,4}
foo<double>(0.1,0.2,0.3) returns std::vector<double> x{0.1,0.2,0.3}

So I suppose you can simply write

template <typename T, typename ... ARGS>
std::vector<T> foo (ARGS const & ... args)
 { return { args... }; }

You can also skip the vector entirely: `for (auto x: { Is... })` — Henri Menke, Nov 05 '17 at 20:43

T33C · Answer 2 · 2017-11-05T23:02:56.157

In foo, I create a vector with the exact capacity to hold all the parameters and then recursively handle each parameter by pushing it to the back of the vector, which is passed to a helper template function called append_to_vector.

#include<vector>    
namespace test2
{
    template<typename TN>
    void append_to_vector(std::vector<TN>& outputvector, const TN& elem)
    {
        outputvector.push_back(elem);
    };

    template<typename T0, typename ...T1toN>
    void append_to_vector(std::vector<T0>& outputvector, const T0& elem, T1toN... elems)
    {
        outputvector.push_back(elem);
        append_to_vector(outputvector, elems...);
    };

    template<typename T, typename ...T0toN>
    auto foo(const T0toN... elems)
    {
        std::vector<T> vec;
        vec.reserve(sizeof...(elems));
        append_to_vector(vec, elems...);
        return vec;
    };
}



int main()
{
   std::vector<int> vec;
   auto vec1 = test2::foo<int>(1,2,3,4);
   auto vec2 = test2::foo<double>(0.1,0.2,0.3,0.4);
   return 0;
}

score 0 · Answer 3 · answered Nov 05 '17 at 23:39

Way to do it in C++14:

#include <vector>
#include <initializer_list>
#include <iterator>
#include <algorithm>
#include <iostream>
template <typename T>
std::vector<T> function(std::initializer_list<T> elements)
{
    return std::vector<T>(elements.begin(), elements.end());
}

int main()
{
    std::vector<int> elements = function<int>({1, 2, 3});
    std::copy(elements.begin(), elements.end(), std::ostream_iterator<const int>(std::cout, " "));
}

converting parameter pack into a vector

3 Answers3

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