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Why does is_constructible claim something is constructible when it isn’t?2个
源自 this CodeReview主题：
源自 this CodeReview主题：
#include <cstddef>
#include <algorithm>
#include <iostream>
#include <type_traits>
#include <utility>
template <typename T>
class aggregate_wrapper : public T {
private:
using base = T;
public:
using aggregate_type = T;
template <typename... Ts>
aggregate_wrapper(Ts&&... xs)
: base{std::forward<Ts>(xs)...} {
// nop
}
};
struct foo_t {
foo_t(int) {}
};
int main() {
std::cout << std::is_constructible<foo_t>::value << std::endl;
std::cout << std::is_constructible<aggregate_wrapper<foo_t>>::value << std::endl;
// aggregate_wrapper<foo_t> v; // won't compile
}
当aggregate_wrapper< foo_t>时,std :: is_constructible< aggregate_wrapper< foo_t>> :: value怎么可能为真v;实际上没有编译？
解决方法
在C标准中,在is_constructible描述中,有一个看似无辜的引用：
Only the validity of the immediate context of the [imaginary
v] variable initialization is considered.
然后说明它意味着什么：
The evaluation of the initialization can result in side
effects such as the instantiation of class template specializations and function template specializations,the
generation of implicitly-defined functions,and so on. Such side effects are not in the immediate context and can result in the program being ill-formed.
我的解释是,当你写：
aggregate_wrapper<foo_t> v;
您正在使用aggregate_wrapper的默认构造函数,它是存在的并且是可访问的,因此它成功,至少在直接上下文中.然后,非立即上下文包括构造函数的主体,并且失败,但这不会改变is_constructible的结果.
