C++11 C++；11右值问题

我不知道下面的示例输出是如何产生的，有人能告诉我吗？提前谢谢

#include <iostream>
#include <algorithm>

class A
{
public:

    // Simple constructor that initializes the resource.
    explicit A(size_t length)
        : mLength(length), mData(new int[length])
    {
        std::cout << "A(size_t). length = "
    << mLength << "." << std::endl;
    }

    // Destructor.
    ~A()
    {
  std::cout << "~A(). length = " << mLength << ".";

  if (mData != NULL) {
            std::cout << " Deleting resource.";
      delete[] mData;  // Delete the resource.
  }

  std::cout << std::endl;
    }

    // Copy constructor.
    A(const A& other)
      : mLength(other.mLength), mData(new int[other.mLength])
    {
  std::cout << "A(const A&). length = "
    << other.mLength << ". Copying resource." << std::endl;

  std::copy(other.mData, other.mData + mLength, mData);
    }

    // Copy assignment operator.
    A& operator=(const A& other)
    {
  std::cout << "operator=(const A&). length = "
           << other.mLength << ". Copying resource." << std::endl;

  if (this != &other) {
      delete[] mData;  // Free the existing resource.
      mLength = other.mLength;
            mData = new int[mLength];
            std::copy(other.mData, other.mData + mLength, mData);
  }
  return *this;
    }

    // Move constructor.
    A(A&& other) : mData(NULL), mLength(0)
    {
        std::cout << "A(A&&). length = " 
             << other.mLength << ". Moving resource.\n";

        // Copy the data pointer and its length from the 
        // source object.
        mData = other.mData;
        mLength = other.mLength;

        // Release the data pointer from the source object so that
        // the destructor does not free the memory multiple times.
        other.mData = NULL;
        other.mLength = 0;
    }

    // Move assignment operator.
    A& operator=(A&& other)
    {
        std::cout << "operator=(A&&). length = " 
             << other.mLength << "." << std::endl;

        if (this != &other) {
          // Free the existing resource.
          delete[] mData;

          // Copy the data pointer and its length from the 
          // source object.
          mData = other.mData;
          mLength = other.mLength;

          // Release the data pointer from the source object so that
          // the destructor does not free the memory multiple times.
          other.mData = NULL;
          other.mLength = 0;
       }
       return *this;
    }

    // Retrieves the length of the data resource.
    size_t Length() const
    {
        return mLength;
    }

private:
    size_t mLength; // The length of the resource.
    int* mData;     // The resource.
};

#include <vector>

int main()
{
   // Create a vector object and add a few elements to it.
   std::vector<A> v;
   v.push_back(A(25));
   v.push_back(A(75));

   ::std::cout << "----------------------" << std::endl;

   // Insert a new element into the second position of the vector.
   //v.insert(v.begin() + 1, A(50));
   return 0;
}

---

std:：vector

在容量耗尽时调整大小时，必须移动或复制其内部状态。但是，它不知道移动它存储的

A

（对

push_back

的两个调用）是否安全，因为移动操作没有标记为

noexcept

。因为

std:：vector

不知道它是安全的，所以它在谨慎和复制值方面犯了错误

// Move constructor.
 A(A&& other) noexcept : mData(NULL), mLength(0) { /* ... */ }

// Move assignment operator.
A& operator=(A&& other) noexcept { /* ... */ }

通过这些更改，我得到以下输出：

A（尺码）。长度=25

.
A(A&&). length = 25. Moving resource.
~A(). length = 0.
A(size_t). length = 75.
A(A&&). length = 75. Moving resource.
A(A&&). length = 25. Moving resource.
~A(). length = 0.
~A(). length = 0.
----------------------
~A(). length = 25. Deleting resource.
~A(). length = 75. Deleting resource.

---

您也可以保留足够的容量（

v.reserve（2）；

在调用

push_-back

）之前，但是如果您的

向量

必须调整大小，您将再次开始复制

A

。

std:：vector

必须在容量耗尽时在调整大小的同时移动或复制其内部状态。但是，它不知道移动它存储的

A

（对

push_back

的两个调用）是否安全，因为移动操作没有标记为

noexcept

。因为

std:：vector

不知道它是安全的，所以它在谨慎和复制值方面犯了错误

// Move constructor.
 A(A&& other) noexcept : mData(NULL), mLength(0) { /* ... */ }

// Move assignment operator.
A& operator=(A&& other) noexcept { /* ... */ }

通过这些更改，我得到以下输出：

A（尺码）。长度=25

.
A(A&&). length = 25. Moving resource.
~A(). length = 0.
A(size_t). length = 75.
A(A&&). length = 75. Moving resource.
A(A&&). length = 25. Moving resource.
~A(). length = 0.
~A(). length = 0.
----------------------
~A(). length = 25. Deleting resource.
~A(). length = 75. Deleting resource.

---

您也可以保留足够的容量（

v.reserve（2）；

在调用

push_-back

）之前，但是如果您的

vector

需要调整大小，您将再次开始复制

A

。

您知道

std:：vector

正在动态分配内存，当它扩展它的内存时，它需要复制它已经拥有的所有数据？请告诉我们你不理解的部分和/或你期望什么。按照@Someprogrammerdude所说的，为了让

vector

在向量重新分配时使用移动操作而不是复制操作，必须将移动构造函数/赋值运算符标记为

noexcept

。请看，您确实知道

std:：vector

正在动态分配内存，并且当它扩展内存时，它需要复制它已经拥有的所有数据？请告诉我们您不理解的部分和/或您期望的内容。按照@Someprogrammerdude所说的，为了使向量重新分配时使用移动操作而不是复制操作，必须将移动构造函数/赋值运算符标记为

noexcept

。明白了！谢谢知道了！谢谢