c ++ 11为什么这个移动构造函数不工作？ [英] c++11 Why doesn't this move constructor work?

问题描述

我写了下面的代码。我希望在 LargeClass 的实例之间移动向量的内容。移动构造函数正在使用，但不是移动我只获得副本。

为什么移动语义在这里不起作用？

代码：

  #include< iostream& 
 #include< vector> 
 
 class LargeClass 
 {
 public：
 explicit LargeClass（void）：numbers（20，10）
 {
} 
 LargeClass（const LargeClass& rhs）：numbers（rhs.numbers）
 {
 std :: cout< Using LargeClass copy constructor<< '\\\
'; 
} 
 LargeClass（const LargeClass&& rhs）：numbers（std :: move（rhs.numbers））
 {
 std :: cout< Using LargeClass move constructor<< '\\\
'; 
} 
 
 const int * getNumbersAddress（void）const 
 {
 return（numbers.data（））; 
} 
 
 private：
 std :: vector< int>数字; 
}; 
 
 int main（）
 {
 LargeClass l1; 
 std :: cout<< l1 vector address：<< l1.getNumbersAddress（）<< '\\\
'; 
 
 LargeClass l2（l1）; 
 std :: cout<< l1 vector address：<< l1.getNumbersAddress（）<< '\\\
'; 
 std :: cout<< l2 vector address：<< l2.getNumbersAddress（）<< '\\\
'; 
 
 LargeClass l3 = std :: move（l2）; 
 std :: cout<< l1 vector address：<< l1.getNumbersAddress（）<< '\\\
'; 
 std :: cout<< l2 vector address：<< l2.getNumbersAddress（）<< '\\\
'; 
 std :: cout<< l3 vector address：<< l3.getNumbersAddress（）<< '\\\
'; 
 
 return 0; 
} 
  

可能的输出：

  l1向量地址：0x18ce010 
使用LargeClass复制构造函数
 l1向量地址：0x18ce010 
 l2向量地址：0x18ce070 
使用LargeClass移动构造函数
 l1矢量地址：0x18ce010 
 l2矢量地址：0x18ce070 
 l3矢量地址：0x18ce0d0 
  

解决方案

右值引用在它们的 const 表单中没有意义，因为你想修改它们到移动）。在C ++中创建为 const 的对象在只读存储器中，从中无法获取/修改内部资源。

移动构造函数的语法 通常应为

• class_name（class_name&&）






所以使用：





LargeClass（LargeClass&& rhs）




I've written the code posted below. I was hoping to get to move the content of a vector between instances of LargeClass. The move constructor is being used, but instead of moving I get copies only.

Why doesn't the move semantics work as expected here?

Code:

#include <iostream>
#include <vector>

class LargeClass
{
public:
    explicit LargeClass (void): numbers(20, 10) 
    {
    }
    LargeClass (const LargeClass &rhs): numbers(rhs.numbers)
    { 
        std::cout << "Using LargeClass copy constructor" << '\n';
    }
    LargeClass (const LargeClass &&rhs): numbers(std::move(rhs.numbers))
    { 
        std::cout << "Using LargeClass move constructor" << '\n';
    }

    const int* getNumbersAddress(void) const
    {
        return (numbers.data());
    }

private:
    std::vector<int> numbers;
};

int main()
{
    LargeClass l1;
    std::cout << "l1 vector address: " << l1.getNumbersAddress() << '\n';

    LargeClass l2(l1);
    std::cout << "l1 vector address: " << l1.getNumbersAddress() << '\n';
    std::cout << "l2 vector address: " << l2.getNumbersAddress() << '\n';

    LargeClass l3 = std::move(l2); 
    std::cout << "l1 vector address: " << l1.getNumbersAddress() << '\n';
    std::cout << "l2 vector address: " << l2.getNumbersAddress() << '\n';
    std::cout << "l3 vector address: " << l3.getNumbersAddress() << '\n';

    return 0;
}

Possible output:

l1 vector address: 0x18ce010
Using LargeClass copy constructor
l1 vector address: 0x18ce010
l2 vector address: 0x18ce070
Using LargeClass move constructor
l1 vector address: 0x18ce010
l2 vector address: 0x18ce070
l3 vector address: 0x18ce0d0

解决方案

The rvalue references don't make sense in their const forms because you want to modify them (you want to "move" them ). Objects created as const in C++ are in read-only memory, from which grabbing/modifying internal resources won't be possible.

The syntax for move constructor normally should be

• class_name ( class_name && )

So use :

LargeClass ( LargeClass&& rhs )

这篇关于c ++ 11为什么这个移动构造函数不工作？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！