移动构造函数(c++常问问题十六)

Posted 2020-06-13

今天我们来讲讲c++11中引入了两个新东西

1.move constructor(移动构造函数)

2.move assignment(移动赋值)

 

Rule of three现在变成了Rule of five（多了上面说的两个东东）

class rule_of_five
{
    char* cstring; // raw pointer used as a handle to a dynamically-allocated memory block
 public:
    rule_of_five(const char* arg)
    : cstring(new char[std::strlen(arg)+1]) // allocate
    {
        std::strcpy(cstring, arg); // populate
    }
    ~rule_of_five()
    {
        delete[] cstring;  // deallocate
    }
    rule_of_five(const rule_of_five& other) // copy constructor
    {
        cstring = new char[std::strlen(other.cstring) + 1];
        std::strcpy(cstring, other.cstring);
    }
    rule_of_five(rule_of_five&& other) : cstring(other.cstring) // move constructor
    {
        other.cstring = nullptr;
    }
    rule_of_five& operator=(const rule_of_five& other) // copy assignment
    {
        char* tmp_cstring = new char[std::strlen(other.cstring) + 1];
        std::strcpy(tmp_cstring, other.cstring);
        delete[] cstring;
        cstring = tmp_cstring;
        return *this;
    }
    rule_of_five& operator=(rule_of_five&& other) // move assignment
    {
        delete[] cstring;
        cstring = other.cstring;
        other.cstring = nullptr;
        return *this;
    }

 

那么新加入的移动赋值以及移动拷贝要怎么使用呢，直接看代码

#include <iostream>
#include <utility>
#include <vector>
#include <string>
int main()
{
    std::string str = "Hello";
    std::vector<std::string> v;
 
    // uses the push_back(const T&) overload, which means 
    // we‘ll incur the cost of copying str
    v.push_back(str);
    std::cout << "After copy, str is \"" << str << "\"\n";
 
    // uses the rvalue reference push_back(T&&) overload, 
    // which means no strings will copied; instead, the contents
    // of str will be moved into the vector.  This is less
    // expensive, but also means str might now be empty.
    v.push_back(std::move(str));
    std::cout << "After move, str is \"" << str << "\"\n";
 
    std::cout << "The contents of the vector are \"" << v[0]
                                         << "\", \"" << v[1] << "\"\n";
}

Output:

After copy, str is "Hello"
After move, str is ""
The contents of the vector are "Hello", "Hello"

看完大概明白一点儿了,加上move之后,str对象里面的内容被"移动"到新的对象中并插入到数组之中了,同时str被清空了。这样一来省去了对象拷贝的过程。所以说在str对象不再使用的情况下，这种做法的效率更高一些！