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

Question

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.

Bornu

You can implement your own make_array() function. – Geiger

Thank you, that is helpful, I might use it. But when such a horribly complex (to me) function is required to do something C could do 30 years, it ago strikes me as wrong somehow. Furthermore I'm concerned my compiler won't be smart enough to compile it to exactly the same as putting the size in myself. – Bornu

C++17 introduced deduction guides for this kind of things, but you have to omit the template arguments. – Papuan

In case someone is looking for a way to initialize a modifiable buffer that is initialized with a string literal without explicitly specifying its size, feel free to check out my answer here: https://mcmap.net/q/158356/-how-to-emulate-c-array-initialization-quot-int-arr-e1-e2-e3-quot-behaviour-with-std-array – Honkytonk

In C++20, there is also: auto a3 = std::to_array<int>({3, 2, 1}); – Trench

Answer 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)...};
}

Devaluation

Yes but can you use that to store std::vector<std::array<type, different_sizes>>? Making a vector of std::array would require all the sizes to be the same? – Storehouse

@Storehouse Unfortunately I think they would need to be the same size. Forget my suggestion in the other thread :( – Bornu

@Storehouse std::vector<std::array<type, different_sizes> > AFAIK is impossible. – Ium

Is there any reason return type deduction can't be used in that last sample or just return {std::forward<T>(values)...}? – Geiger

Would @Konrad Rudolph's simplified version of make_array() above be better than the Intermediate version (in updated original question) posted here ? What are the pro's and cons of each ? – Kennie

The final example is very nice, except that I don't understand why it repeats the std::array<> type when declaring it via auto -> enables us simply to return { the_initializer_list } without repeating the typename. (Or this certainly compiles fine for me, anyway., just like normal) – Prasad

Answer 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)... }};
}

Hoogh

Yeah, wow, the 2nd example works perfectly for most cases and isn't complex at all. Thanks! – Prasad

The only difference between the two proposed implementations is the extra set of curly braces? What syntactic difference does the seemingly unnecessary pair of braces do? – Appropriation

To add to these comments (albeit very late), the double braces are related to std::array initialization. This is described here. I think the reason why {std::forward<N>(args)...}; works is the slight relaxation of the rules for eliding braces, described here. – Bertha