在编译时零初始化算术 `value_type` 的 `std::array` 会导致缺少构造函数注释

Zero-initializing a `std::array` of arithmetic `value_type` at compile-time results in a missing constructor note

为了在编译时初始化算术类型 ATstd::array,我这样做了:

#include <array> 
#include <iostream>

template<typename AT, auto DIM, auto N = 0> 
constexpr void assign(std::array<AT, DIM>& arr, AT value)
{
    arr[N] = value; 
    if constexpr (N < std::size(arr) - 1)
        assign<AT, DIM, N + 1>(arr, value);
}

template<typename AT, auto DIM>
constexpr std::array<AT, DIM> zero_array() 
{
    std::array<AT, DIM> result;  
    assign(result, AT{0});
    return result;
}

template<typename Container> 
void print(Container&& cont)
{
    for (const auto& el : cont)
        std::cout << el << " "; 
    std::cout << std::endl;
}

int main()
{
    auto zero10 = zero_array<double, 10>(); 
    print(zero10); 
}

godbolted it,在我看来它有效:

然而,当我用

编译它时
g++  -O3 -std=c++2a -Wall -Wpedantic -Wunused-parameter -I /usr/include main.cpp -o main

使用 g++ (GCC) 8.2.1 20181127,我得到 "note"

In file included from main.cpp:1:
main.cpp: In instantiation of ‘constexpr std::array<AT, DIM> zero_array() [with AT = double; auto DIM = 10]’:
main.cpp:30:42:   required from here
/usr/include/c++/8.2.1/array:94:12: note: ‘struct std::array<double, 10>’ has no user-provided default constructor
     struct array
            ^~~~~
/usr/include/c++/8.2.1/array:110:56: note: and the implicitly-defined constructor does not initialize ‘double std::array<double, 10>::_M_elems [10]’
       typename _AT_Type::_Type                         _M_elems;

为什么会有这张便条?我可以忽略它吗?如果是这样,如何摆脱它?

std::array 没有默认构造函数。您需要使用:

std::array<AT, DIM> result = {};

令我困惑的是为什么你认为你需要 zero_array。如果你使用

std::array<double, 10> a = {};

你得到一个零初始化的对象。