初始化非默认可构造元素的 std::array?

Initializing an std::array of non-default-constructible elements?(初始化非默认可构造元素的 std::array?)

问题描述

假设类型 foo_t 具有命名构造函数习惯用法 make_foo().现在，我想要正好有 123 个 foo ——不多也不少.所以，我正在考虑一个 std::array<foo_t, 123>.现在，如果 foo_t 是默认可构造的，我会写:

Suppose type foo_t with a named constructor idiom, make_foo(). Now, I want to have exactly 123 foo's - no more, no less. So, I'm thinking about an std::array<foo_t, 123>. Now, if foo_t were default-constructible, I would write:

std::array<foo_t, 123> pity_the_foos;
std::generate(
    std::begin(pity_the_foos), std::end(pity_the_foos),
    []() { return make_foo(); }
);

Bob 是我的叔叔，对吧?不幸的是... foo_t 没有默认 ctor.

and Bob's my uncle, right? Unfortunately... foo_t has no default ctor.

那么我应该如何初始化我的数组呢?我是否需要使用一些可变参数模板扩展巫术?

How should I initialize my array, then? Do I need to use some variadic template expansion voodoo perhaps?

注意:如果有帮助，答案可以使用 C++11、C++14 或 C++17 中的任何内容.

Note: Answers may use anything in C++11, C++14 or C++17 if that helps at all.

推荐答案

平常.

template<size_t...Is>
std::array<foo_t, sizeof...(Is)> make_foos(std::index_sequence<Is...>) {
    return { ((void)Is, make_foo())... };
}

template<size_t N>
std::array<foo_t, N> make_foos() {
    return make_foos(std::make_index_sequence<N>());
}

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