Is there a way to enforce full initialization of std::array

You can enforce this by using parameter pack, but with a bit different syntax:

#include <array>template<size_t N>struct A{   template<typename... T>   size_t function(T&&... nums)   {     static_assert(sizeof...(nums) == N, "Wrong number of arguments");     std::array<size_t, N> arr = { std::forward<size_t>(nums)... };     return arr[N-1];   }};int main(){   A<5> a;   a.function(1,2,3,4,5); // OK   a.function(1,2,4,5);   // Compile-time error}

But, I think there is no good way to enforce that in compile time. I would just use assert(il.size() == N) in production code to check size of initializer list.

You can create a wrapper for your objects

template <class T>struct WrapT{    WrapT() = delete;    WrapT(T e)   : value(e){}   public: T value;    // or add some operator()};

and

size_t function(std::array<WrapT<size_t>, N> arr){ return arr[N-1].value;}

so a function call like (with full brace-init)

function({ {{1}, {2}, {3}, {4}} }); 

will not compile because of use of deleted function. Live example. The syntax is a little clumsy, and even then I'm not sure all the possible value-initialization cases are covered.

@dpy points out you may omit the innermost ones, so you get to your original code: function( {{ 1, 2, 3, 4, 5 }} ).

Simple answer: You cannot.

When initializing std::array with a list, it is doing an aggregate initialization, and it is explained here when the list size is less than the number of member:

• If the number of initializer clauses is less than the number of members or initializer list is completely empty, the remaining members are initialized by their brace-or-equal initializers, if provided in the class definition, and otherwise (since C++14) by empty lists, which performs value-initialization. If a member of a reference type is one of these remaining members, the program is ill-formed (references cannot be value-initialized)

It is simply a legal and acceptable behavior to provide less than the size list, so compiler will not complain anything. Your code:

A<5> a;a.function({{1,2,3}}));

is equivalent to:

A<5> a;a.function({{1,2,3,0,0}}));

to compiler. Your best bet is runtime error (which may not be as you wished).