[英]Initializing an std::array of non-default-constructible elements?
Suppose type foo_t
with a named constructor idiom, make_foo()
.假设类型
foo_t
具有命名构造函数, make_foo()
。 Now, I want to have exactly 123 foo's - no more, no less.现在,我想要正好有 123 个 foo ——不多也不少。 So, I'm thinking about an
std::array<foo_t, 123>
.所以,我正在考虑一个
std::array<foo_t, 123>
。 Now, if foo_t
were default-constructible, I would write:现在,如果
foo_t
是默认可构造的,我会写:
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?鲍勃是我叔叔,对吧? Unfortunately...
foo_t
has no default ctor.不幸的是......
foo_t
没有默认的ctor。
How should I initialize my array, then?那我应该如何初始化我的数组呢? Do I need to use some variadic template expansion voodoo perhaps?
我是否需要使用一些可变参数模板扩展巫术?
Note: Answers may use anything in C++11, C++14 or C++17 if that helps at all.注意:如果有帮助,答案可能会使用 C++11、C++14 或 C++17 中的任何内容。
The usual.通常。
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>());
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.