如何STL容器复制对象？ [英] How do STL containers copy objects?

问题描述

我知道STL容器如向量在添加对象时复制对象。 push_back 方法如下：

  void push_back（const T& x） ; 
  

我惊讶地发现它将该项目作为引用。我写了一个示例程序来看看它是如何工作的。

  struct Foo 
 {
 Foo b $ b {
 std :: cout<< Inside Foo constructor< std :: endl; 
} 
 
 Foo（const Foo& f）
 {
 std :: cout< inside copy constructor< std :: endl; 
} 
}; 
 
 Foo f; 
 std :: vector< Foo> foos; 
 foos.push_back（f）; 
  

这样就复制了对象，我可以看到它正在调用 copy-constructor 。

我的问题是，当 push_back 将项目作为引用时，它如何调用复制构造函数？

解决方案

任何想法...

p>它可能使用placement new 在其内部数组中就地构造对象。放置 new 不分配任何内存;它只是放置你指定的对象，并调用构造函数。语法是 new（address）Class（constructor_arguments）。

复制构造函数 T :: T（T const&）被调用以就地创建副本。类似这样的东西（简化）：

 模板< T& 
 void vector< T> :: push_back（T const& item）{
 //如果必要，调整大小
 new（& d_array [d_size ++]）T 
} 
  

注意 T 必须有一个复制构造函数为此工作。默认情况下（如果你什么都不做），它会免费获得一个。如果您明确定义它，则向量< T> 必须 public 才能工作。

这是GNU的libstdc ++的功能它，但我怀疑这将是非常有启发性。有一个分配器（向量的第二个模板参数），这使它不那么直接。

I know STL containers like vector copies the object when it is added. push_back method looks like:

void push_back ( const T& x );

I am surprised to see that it takes the item as reference. I wrote a sample program to see how it works.

struct Foo
{
    Foo()
    {
        std::cout << "Inside Foo constructor" << std::endl;
    }

    Foo(const Foo& f)
    {
        std::cout << "inside copy constructor" << std::endl;
    }
};

Foo f;
std::vector<Foo> foos;
foos.push_back(f);

This copies the object and I can see it is calling copy-constructor.

My question is, when the push_back takes item as reference, how it is calling copy-constructor? Or am I missing something here?

Any thoughts..?

解决方案

It probably uses "placement new" to construct the object in-place in its internal array. Placement new doesn't allocate any memory; it just places the object where you specify, and calls the constructor. The syntax is new (address) Class(constructor_arguments).

The copy constructor T::T(T const &) is called to create the copy in-place. Something like this (simplified):

template<T>
void vector<T>::push_back(T const &item) {
    // resize if necessary
    new (&d_array[d_size++]) T(item);
}

Note that T must have a copy constructor for this to work. By default (if you do nothing), it gets one for free. If you define it explicitly, it must be public for vector<T> to work.

Here's how GNU's libstdc++ does it, but I doubt that it'll be very enlightening. There is an allocator (the second template argument to vector) that makes it less straightforward.

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