How to create std::array with initialization list without providing size directly [duplicate]

问题

This question already has an answer here:

How to emulate C array initialization “int arr[] = { e1, e2, e3, … }” behaviour with std::array? 9 answers

How can I make a3 compile?

int main()
{
    int a1[] = { 1, 2, 3 };
    std::array<int, 3> a2 = { 1, 2, 3 };
    std::array<int> a3 = { 1, 2, 3 };
}

It's very inconvenient, and brittle, to hard-code the size of the array when using an initialization list, especially long ones. Is there any work around? I hope so otherwise I'm disappointed because I hate C arrays and std::array is supposed to be their replacement.

回答1:

There is currently no way to do this without rolling your own make_array, there is a proposal for this N3824: make_array which has the following scope:

LWG 851 intended to provide a replacement syntax to
array<T, N> a = { E1, E2, ... };
, so the following
auto a = make_array(42u, 3.14);
is well-formed (with additional static_casts applied inside) because
array<double, 2> = { 42u, 3.14 };
is well-formed.

This paper intends to provide a set of std::array creation interfaces which are comprehensive from both tuple’s point of view and array’s point of view, so narrowing is just naturally banned. See more details driven by this direction in Design Decisions.

It also includes a sample implementation, which is rather long so copying here is impractical but Konrad Rudolph has a simplified version here which is consistent with the sample implementation above:

template <typename... T>
constexpr auto make_array(T&&... values) ->
    std::array<
       typename std::decay<
           typename std::common_type<T...>::type>::type,
       sizeof...(T)> {
    return std::array<
        typename std::decay<
            typename std::common_type<T...>::type>::type,
        sizeof...(T)>{std::forward<T>(values)...};
}

回答2:

You're being a little overdramatic when you say "such a horribly complex (to me) function is required". You can make a simplified version yourself, the proposal also includes a "to_array" function to convert C-arrays and deducing the type from the first parameter. If you leave that out it gets quite manageable.

template<typename T, typename... N>
auto my_make_array(N&&... args) -> std::array<T,sizeof...(args)>
{
    return {std::forward<N>(args)...};
}

which you can then call like

auto arr = my_make_array<int>(1,2,3,4,5);

edit: I should mention that there actually is a version of that in the proposal that I overlooked, so this should be more correct than my version:

template <typename V, typename... T>
constexpr auto array_of(T&&... t)
    -> std::array < V, sizeof...(T) >
{
    return {{ std::forward<T>(t)... }};
}

来源：https://stackoverflow.com/questions/26351587/how-to-create-stdarray-with-initialization-list-without-providing-size-directl

标签

c++

initialization

stdarray