How to list-initialize a const std::array member using a std::array constructor argument C++

Question

suppose we have the following class in C++11 or later:

class MyClass {
private:
    const std::array<SomeType, 100> myArray;
public:
    explicit MyClass(std::array<SomeOtherType, 100> initArray);
};

Assuming that class SomeType has a constructor that takes a single SomeOtherType as an argument, is it possible to initialize the const member array using list-initialization in the constructor? What is the syntax for doing so?

Clearly, just directly initializing it like this doesn't work:

MyClass::MyClass(std::array<SomeOtherType, 100> initArray) :
    myArray{initArray} {}

Thanks!

You are aware that your class with a const member won't be copyable? — Swordfish
– Swordfish, Commented Nov 27, 2018 at 3:55
if you are using c++11 (or newer), look up documentation for std::initializer_list. If you are compiling with GCC, you may also find this post useful stackoverflow.com/questions/8192185/… — HappyKeyboard
– HappyKeyboard, Commented Nov 27, 2018 at 3:58
@Swordfish: Thanks for the heads up - that shouldn't be a problem for my current use case, but I'll keep that in mind for the future. — Embedder
– Embedder, Commented Nov 27, 2018 at 4:37
@HappyKeyboard: Thanks, I'll take a look at std::initializer_list. Also good to know about the need for double curly brackets with std::array and gcc! — Embedder
– Embedder, Commented Nov 27, 2018 at 4:39

Swordfish · Accepted Answer · 2018-11-27 12:13:36Z

You can use a variadic template:

#include <array>

struct foo
{
    const std::array<int, 10> bar;

    template<typename... T>
    foo(T&&... t)
    : bar({ std::move(t)... })
    {}
};

int main()
{
    foo f{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
}

Or you can initialize it with an array passed to the constructor:

#include <array>

struct foo
{
    const std::array<int, 10> bar;

    explicit foo(std::array<int, 10> const &qux)
    : bar{ qux }
    {}
};

int main()
{
    std::array<int, 10> qux;
    foo f(qux);
}

But these options don't take into account that you want to have an array of SomeOtherType converted to an array of SomeType. I didn't realize that at first, hece the variants above.

#include <cstddef>
#include <array>
#include <utility>

struct SomeOtherType{};

struct SomeType {
    SomeType(SomeOtherType) {}
};

struct MyClass
{
    const std::array<SomeType, 100> myArray;

    template<typename T, std::size_t... N>
    MyClass(T&& qux, std::index_sequence<N...>)
    : myArray{ qux[N]... }
    {}

    explicit MyClass(std::array<SomeOtherType, 100> const &qux)
    : MyClass{ qux, std::make_index_sequence<100>{} }
    {}
};

int main()
{
    std::array<SomeOtherType, 100> qux{};
    MyClass foo(qux);
}

Passer By · Accepted Answer · 2018-11-27 04:17:44Z

1

You can unpack arguments with std::index_sequence and delegating constructors

template<typename Arr, size_t... Is>
MyClass(Arr&& arr, std::index_sequence<Is...>)
  : myArray{arr[Is]...} ()

explicit MyClass(std::array<SomeOtherType, 100> arr) : MyClass(arr, std::make_index_sequence<100>{}) ()

answered Nov 27, 2018 at 4:17

Passer By

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Comments

Peter Ruderman · Accepted Answer · 2018-11-27 04:18:33Z

0

This is possible. You just need a little helper function template to do the conversion for you. Something like this:

template <class T, class U, size_t N>
std::array<T, N> ArrayConvert(std::array<U, N> const& init)
{
  std::array<T, N> result;
  std::copy(init.begin(), init.end(), result.begin());
  return result;
}

class Foo
{
  std::array<int, 100> myArray;
public:
  template <class U> Foo(std::array<U, 100> const& init)
    : myArray(ArrayConvert<int>(init))
  {
  }
};

answered Nov 27, 2018 at 4:18

Peter Ruderman

12.6k1 gold badge41 silver badges59 bronze badges

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