Initializing an std::array of non-default-construc

2020-03-01 18:22发布

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(); }
);

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

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

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

1条回答
兄弟一词,经得起流年.
2楼-- · 2020-03-01 19:01

The usual.

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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