移動建構函式(c++常問問題十六)

標籤：

今天我們來講講c++11中引入了兩個新東西

1.move constructor(移動建構函式)

2.move assignment(移動賦值)

 

Rule of three現在變成了Rule of five（多了上面說的兩個東東）

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;    }

 

那麼新加入的移動賦值以及移動拷貝要怎麼使用呢，直接看代碼

#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"

看完大概明白一點兒了,加上move之後,str對象裡面的內容被"移動"到新的對象中並插入到數組之中了,同時str被清空了。這樣一來省去了對象拷貝的過程。所以說在str對象不再使用的情況下，這種做法的效率更高一些！

移動建構函式(c++常問問題十六)