为什么用C ++ 11更改了std :: vector的构造函数接口？ [英] Why was the constructor interface of std::vector changed with C++11?

问题描述

为什么新标准中删除了默认参数？我经常构造一个矢量变量，如下所示： std :: vector< my_pod_struct> buf（100）。我想我会在C ++ 11编译器中遇到编译器错误。

 显式矢量（size_type count，
 const T& value = T（），/ *直到C ++ 11 * / 
 const Allocator& alloc = Allocator（））; 
 vector（size_type count，
 const T&值，/ *自C ++ 11起* / 
 const Allocator& alloc = Allocator（））; 
  

解决方案

之前，当您写 std :: vector< T> buf（100）; ，您将获得一个默认构造的 T ，然后将该实例复制到向量中的一百个插槽中。 / p>

现在，当您编写 std :: vector< T> buf（100）; ，它将使用另一个构造函数： explicit vector（size_type count）; 。这将默认构造一百个 T s。

新的单参数构造函数不需要类型 T 是可复制的。这很重要，因为现在类型可以移动并且不能复制。

Why was the default argument removed with the new standard? Often I constructed a vector variable like this: std::vector<my_pod_struct> buf(100). I guess I would get an compiler error with a C++11 compiler.

explicit vector( size_type count,
                 const T& value = T(),                   /* until C++11 */
                 const Allocator& alloc = Allocator());
         vector( size_type count,
                 const T& value,                         /* since C++11 */
                 const Allocator& alloc = Allocator());

解决方案

Before, when you wrote std::vector<T> buf(100); you would get one T default constructed, and then that instance would be copied over to one hundred slots in the vector.

Now, when you write std::vector<T> buf(100);, it will use another constructor: explicit vector( size_type count );. This will default-construct one hundred Ts. It's a slight difference, but an important one.

The new single-argument constructor doesn't require the type T to be copyable. This is important because now types can be movable and not copyable.

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