std::initializer list from already existing std::array without enumerating each element

Question

Lets assume that I have a class with the following constructor:

class Foo {
    Foo(std::initializer_list<uin8_t> args)
    {
     ....
    }
}

and I have the following array: std::array<uint8_t, 6> bar.

Now I would like to create an object off foo with the bar array. Is there an other way as doing it in the follwing way:

Foo f (bar[0], bar[1], bar[2], bar[3], bar[4], bar[5]);

This way seems a bit complicated and it feels like this is not the way it should be.

So, can I create a std::initializer list from an already existing array, without enumerating each array element?

Why not just add a overload to the constructor set that takes a std::array? — NathanOliver
– NathanOliver, Commented Aug 7, 2017 at 12:01
No, it's not possible. Actually you can't explicitly create anything but an empty std::initializer_list, as there are no functionality to add elements to it. — Some programmer dude
– Some programmer dude, Commented Aug 7, 2017 at 12:01
@NathanOliver Thats just because the Foo class comes from an other API so I wanted to know if I can use this constructor in a more effective way. — Ventu
– Ventu, Commented Aug 7, 2017 at 12:06
As for another possible solution, rather an adding an overload for std::array, how about creating a constructor taking a generic range using iterators? Though it only works if you can actually modify the class. — Some programmer dude
– Some programmer dude, Commented Aug 7, 2017 at 12:07
@Someprogrammerdude also a good idea, but in this case I hoped that I do not have to touch the given API. But I will think about it. Thanks — Ventu
– Ventu, Commented Aug 7, 2017 at 12:08

John Zwinck · Accepted Answer · 2017-08-07 12:38:03Z

2

No, you can't do it. This API which only accepts an initializer_list and has no facility to accept, say, a pair of iterators, or a pointer plus a size, is deficient. You'll virtually never see an API like this in the wild.

answered Aug 7, 2017 at 12:38

John Zwinck

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bolov · Accepted Answer · 2017-08-07 12:47:49Z

2

since you cannot modify "Foo" you can create your own make_foo method that automatically expands the array:

struct Foo
{
    Foo(std::initializer_list<int> args) {}
};

template <std::size_t N, std::size_t... Is>
auto make_foo(std::array<int, N>& arr, std::index_sequence<Is...>) -> Foo
{
  return Foo{arr[Is]...};
}


template <std::size_t N>
auto make_foo(std::array<int, N>& arr) -> Foo
{
  return make_foo(arr, std::make_index_sequence<N>{});
}

auto test()
{
  std::array<int, 4> arr = {{1, 2, 3, 4}};

  auto foo = make_foo(arr);
}

answered Aug 7, 2017 at 12:47

bolov

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