How to pass an array to std::make_tuple?

Question

How to do this:

#include <tuple>

std::tuple<double, int[2]> f()
{
    return std::make_tuple(0., {1, 2});
}

int main()
{
    f();
    return 0;
}

could not convert 'std::make_tuple<{}>()' from 'std::tuple<>' to 'std::tuple'

Can't the compiler figure out what to do?

score 5 · Answer 1 · answered Dec 12 '16 at 14:48

If you necessary need to use a "compile-time vector", try array.

#include <array>
// ...

std::tuple<double, std::array<int, 2>> f() {
  return std::make_tuple(0., std::array<int, 2>{1, 2});
}

As an aggregate type, it can be initialized with aggregate-initialization given at most N initializers that are convertible to T.

score 2 · Answer 2 · edited May 23 '17 at 12:13

2

Instead, you may choose to pass a vector like this:

std::tuple<double, vector<int>> t = std::make_tuple(0.0, vector<int>{1,3});

edited May 23 '17 at 12:13

Community

answered Dec 12 '16 at 14:30

Saurav Sahu

@Piotr Skotnicki, that "int pointer" example made me even verify with `sizeof` that `int[2]` doesn't decay to a ponter in the tuple. Anyway, for the practical application I'll go with `using MyTuple = std::tuple; ... return MyTuple(0., {1, 2});`. – Velkan Dec 12 '16 at 15:08

2 Answers2