c – std :: is_constructible没有给出正确的结果

参见英文答案 > Why does is_constructible claim something is constructible when it isn’t?2个
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.

原文地址：https://www.jb51.cc/c/118493.html