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std :: initializer_list作为构造函数的模板参数

[英]std::initializer_list as template argument for constructor

Consider a class which inherits from a std container with a template constructor which calls the underlying constructor of the container. 考虑一个继承自std容器的类,该类具有调用容器的底层构造函数的模板构造函数。 This template constructor works for the simple copy and move constructor but not for the initializer_list ctor. 此模板构造函数适用于简单复制和移动构造函数,但不适用于initializer_list ctor。

template<typename container_T>
class test : public container_T {
public:
  using container_type = container_T;

  test() {} 

  // templated constructor
  template<typename T>
  test(T t)
    : container_T(t) {}

   // without this it won't compile
  test(std::initializer_list<typename container_T::value_type> l)
    : container_T(l) {}
};

int main() {    
  test<std::deque<int>> vdi1;
  test<std::deque<int>> vdi2({1,2,3,4,5,6,7,8,9});

  std::cout << "vdi2 before:" << std::endl;
  for(auto it : vdi2)
    std::cout << it << std::endl;

  test<std::deque<int>> vdi3(std::move(vdi2));

  std::cout << "vdi2 before:" << std::endl;
  for(auto it : vdi2)
    std::cout << it << std::endl;

  std::cout << "vdi3 before:" << std::endl;
  for(auto it : vdi3)
    std::cout << it << std::endl;

  return 0;
}

If I remove the initializer_list constructor vdi2 won't compile. 如果我删除了initializer_list构造函数, vdi2将无法编译。 So my question: Why is the initializer_list not deduced by the template constructor? 所以我的问题:为什么初始化列表不是由模板构造函数推导出来的? And is it possible to do so? 这样可能吗?

why is the initializer_list not deduced by the templated constructor? 为什么initializer_list不是由模板化的构造函数推导出来的?

The reason is that {1,2,3,4,5,6,7,8,9} is just a synctatic construct that doesn't have a type. 原因是{1,2,3,4,5,6,7,8,9}只是一个没有类型的同构构造。 Therefore, the compiler cannot deduce a type T for this synctatic construct and the first constructor fails. 因此,编译器无法为此synctatic构造推导出类型T ,并且第一个构造函数失败。

However, by special Standard rules std::initializer_list<T> (among other things) can be construct from this synctatic construct and T can be deduced to int . 但是,通过特殊的标准规则, std::initializer_list<T> (以及其他内容)可以从此synctatic构造中构造,并且T可以推导为int Hence the second constructor works. 因此第二个构造函数工作。

By constrast with function template argument type deduction, with 通过与函数模板参数类型推导的约束,用

auto x = {1,2,3,4,5,6,7,8,9};

the compiler sets the type of x to be std::initializer_list<int> . 编译器将x的类型设置为std::initializer_list<int> There are also special Standard rules that says it must be so. 还有一些特殊的标准规则,它必须如此。 Strictly speaking this is not type deduction because, as said above, {1,2,3,4,5,6,7,8,9} doesn't have a type to be deduced. 严格地说,这不是类型推断,因为如上所述, {1,2,3,4,5,6,7,8,9}没有要推断的类型。 (The only type deduction happening here is T = int in std::initializer_list<T> .) Here the compiler chooses (it doesn't deduce) the type of x to be std::initializer_list<int> . (此处发生的唯一类型推导是std::initializer_list<T> T = int 。)这里编译器选择(不推断) x的类型为std::initializer_list<int> In any case, there's no harm to use the abuse of language of saying that the type of x is deduced to std::initializer_list<int> . 在任何情况下,使用滥用语言说x的类型推导到std::initializer_list<int>没有任何害处。

Finally, as DyP said in the comments, what you probably want is inheriting all constructors (not only those taking one argument) from the base container class. 最后,正如DyP在评论中所说,你可能想要的是从基础容器类继承所有构造函数(不仅仅是那些带有一个参数的构造函数)。 You can do this by removing all the constructors that you currently have and add just this line to test : 您可以通过删除当前拥有的所有构造函数来执行此操作,并添加此行以进行test

using container_type::container_type;

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