c++ - 微软示例的这个移动构造函数似乎有冗余代码

Question

MSDN page对于开发人员有这个代码片段:

// Move constructor.  
MemoryBlock(MemoryBlock&& other) : _data(nullptr), _length(0)  
{  
   std::cout << "In MemoryBlock(MemoryBlock&&). length = "   
             << other._length << ". Moving resource." << std::endl;  

   // Copy the data pointer and its length from source object.  
   _data = other._data;      // Assginment 1
   _length = other._length;  // Assignment 2

   // Release the data pointer from the source object so that  
   // the destructor does not free the memory multiple times.  
   other._data = nullptr;  
   other._length = 0;  
}

当标记为Assignment 1 和Assignment 2 的指令时，_data(nullptr)、_length(0) 有什么用 覆盖_data 和_length 的值？

Answer 1

应该是吧

// Move constructor.  
MemoryBlock(MemoryBlock&& other) : _data(other._data), _length(other._length)
{  
   std::cout << "In MemoryBlock(MemoryBlock&&). length = "   
             << other._length << ". Moving resource." << std::endl;  

   // Release the data pointer from the source object so that  
   // the destructor does not free the memory multiple times.  
   other._data = nullptr;  
   other._length = 0;  
}