初始化非默认可构造元素的std :: array吗? [英] Initializing an std::array of non-default-constructible elements?
问题描述
假设类型foo_t
具有命名的构造函数成语make_foo()
.现在,我想精确地拥有123个foo-不多也不少.因此,我正在考虑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(); }
);
鲍勃是我的叔叔,对吧?不幸的是... foo_t
没有默认的ctor.
and Bob's my uncle, right? Unfortunately... foo_t
has no default ctor.
那我应该如何初始化我的数组?我是否需要使用一些可变参数模板扩展伏都教?
How should I initialize my array, then? Do I need to use some variadic template expansion voodoo perhaps?
注意:如果有帮助的话,答案可以在C ++ 11,C ++ 14或C ++ 17中使用.
Note: Answers may use anything in C++11, C++14 or C++17 if that helps at all.
推荐答案
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>());
}
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