How to Construct Std::Array Object with Initializer List

How to construct std::array object with initializer list?

An std::array<> has no constructor that takes an std::initializer_list<> (initializer list constructor) and there is no special language support for what it may mean to pass a std::initializer_list<> to a class' constructors such that that may work. So that fails.

For it to work, your derived class needs to catch all elements and then forward them, a constructor template:

template<typename ...E>
enum_addressable_array(E&&...e) : base_t{{std::forward<E>(e)...}} {}

Note that you need {{...}} in this case because brace elision (omitting braces like in your case) does not work at that place. It's only allowed in declarations of the form T t = { ... }. Because an std::array<> consists of a struct embedding a raw array, that will need two level of braces. Unfortunately, I believe that the exact aggregate structure of std::array<> is unspecified, so you will need to hope that it works on most implementations.

How to initialize std::array member in constructor initialization list when the size of the array is a template parameter

std::index_sequence was provided in order to simplify the metaprogramming task of creating and expanding a pack whose size is not fixed. Use std::make_index_sequence and delegate the construction to some private constructor that deduces the pack:

A(H h) : A(h, std::make_index_sequence<N>{}) {}

template <std::size_t... i>
A(H h, std::index_sequence<i...>) : hashes{((void)i, h)...} {}

Here, ((void)i, h) is a comma expression that evaluates to h, but since we mentioned the pack i inside it, it's eligible to be expanded using .... The result is N copies of h (one for each element of i). This expansion occurs inside a braced-init-list, so the result is a braced-init-list contaning N copies of h, which will then initialize the hashes member.

Initialize an std::array of objects with non-trivial constructor

You might write helper class:

template <std::size_t ... Is, typename T>
std::array<T, sizeof...(Is)> make_array_impl(const T& def, std::index_sequence<Is...>)
{
    return {{(Is, void(), def)...}};
}

template <std::size_t N, typename T>
std::array<T, N> make_array(const T& def)
{
    return make_array_impl(def, std::make_index_sequence<N>());
}

and then

Cube(int size) : c_edgelets(make_array<12>(Edgelet(size - 2)) {}

Using std::array with initialization lists

std::array is funny. It is defined basically like this:

template<typename T, int size>
struct std::array
{
  T a[size];
};

It is a struct which contains an array. It does not have a constructor that takes an initializer list. But std::array is an aggregate by the rules of C++11, and therefore it can be created by aggregate initialization. To aggregate initialize the array inside the struct, you need a second set of curly braces:

std::array<std::string, 2> strings = {{ "a", "b" }};

Note that the standard does suggest that the extra braces can be elided in this case. So it likely is a GCC bug.

Initialization list for an array of objects that that take parameters C++

If you have C++11, you can do:

B(int R) : myA{1, 2, 3, 4, 5, 6} { /* do something */ }

Although, if you are using Visual Studio 2013, please note that this syntax is currently NOT supported. There is the following workaround, however (which is arguably better style anyways):

#include <array>

class B {
    std::array<A, N> myA;

public:
    B(int R) : myA({1, 2, 3, 4, 5, 6}) {}
};

Note, however, that N has to be a compile-time constant, and the number of initializers has to match the number of initializers.

How come std::initializer_list is allowed to not specify size AND be stack allocated at the same time?

The thing is, std::initializer_list does not hold the objects inside itself. When you instantiate it, compiler injects some additional code to create a temporary array on the stack and stores pointers to that array inside the initializer_list. For what its worth, an initializer_list is nothing but a struct with two pointers (or a pointer and a size):

template <class T>
class initializer_list {
private:
  T* begin_;
  T* end_;
public:
  size_t size() const { return end_ - begin_; }
  T const* begin() const { return begin_; }
  T const* end() const { return end_; }

  // ...
};

When you do:

foo({2, 3, 4, 5, 6});

Conceptually, here is what is happening:

int __tmp_arr[5] {2, 3, 4, 5, 6};
foo(std::initializer_list{arr, arr + 5});

One minor difference being, the life-time of the array does not exceed that of the initializer_list.

How do I initialize a member array with an initializer_list?

You can use a variadic template constructor instead of an initializer list constructor:

struct foo { 
    int x[2]; 
    template <typename... T> 
    foo(T... ts) : x{ts...} { // note the use of brace-init-list
    } 
};

int main() {
    foo f1(1,2);   // OK
    foo f2{1,2};   // Also OK
    foo f3(42);    // OK; x[1] zero-initialized
    foo f4(1,2,3); // Error: too many initializers
    foo f5(3.14);  // Error: narrowing conversion not allowed
    foo f6("foo"); // Error: no conversion from const char* to int
}

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EDIT: If you can live without constness, another way would be to skip initialization and fill the array in the function body:

struct foo {
    int x[2]; // or std::array<int, 2> x;
    foo(std::initializer_list<int> il) {
       std::copy(il.begin(), il.end(), x);
       // or std::copy(il.begin(), il.end(), x.begin());
       // or x.fill(il.begin());
    }
}

This way, though, you lose the compile-time bounds checking that the former solution provides.

Can I initialize an array using the std::initializer_list instead of brace-enclosed initializer?

Other answered correctly said this is not possible upfront. But with little helpers, you can get pretty close

template<typename T, std::size_T N, std::size_t ...Ns>
std::array<T, N> make_array_impl(
    std::initializer_list<T> t,
    std::index_sequence<Ns...>) 
{
    return std::array<T, N>{ *(t.begin() + Ns) ... };
}

template<typename T, std::size_t N>
std::array<T, N> make_array(std::initializer_list<T> t) {
    if(N > t.size())
       throw std::out_of_range("that's crazy!");
    return make_array_impl<T, N>(t, std::make_index_sequence<N>());
}

If you are open to more work arounds, you can put this into a class to catch statically-known length violations for the cases where you pass a braced init list. But be warned that most people who read this code will head-desk

template<typename T, std::size_t N>
struct ArrayInitializer {
    template<typename U> struct id { using type = U; };
    std::array<T, N> t;

    template<typename U = std::initializer_list<T>>
    ArrayInitializer(typename id<U>::type z) 
        :ArrayInitializer(z, std::make_index_sequence<N>())
    { 
        if(N > z.size())
            throw std::out_of_range("that's crazy!");
    }

    template<typename ...U>
    ArrayInitializer(U &&... u)
       :t{ std::forward<U>(u)... }
    { }

private:
    template<std::size_t ...Ns>
    ArrayInitializer(std::initializer_list<T>& t,
                     std::index_sequence<Ns...>)
       :t{ *(t.begin() + Ns) ... }
    { }
};

template<typename T, std::size_t N>
std::array<T, N> f(ArrayInitializer<T, N> ai) {
    return std::move(ai.t);
}

int main() {
   f<int, 5>({1, 2, 3, 4, 5});  // OK 
   f<int, 5>({1, 2, 3, 4, 5, 6});  // "too many initializers for array<int, 5>"

   std::initializer_list<int> il{1, 2, 3, 4, 5};
   f<int, 5>(il); // ok
}

Note that both the non-static case at the top of the answer and the "head-desk" case do only check whether you provide too few initializing elements, and errors out then, for the initializer_list case. If you provide too many for the initializer_list case, the trailing elements are just ignored.

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