C++移动语义介绍与使用讲解

class TestClass
{
public:
    TestClass(int s) :m_number(s) {
        cout << "constructor!\n";
    }
    ~TestClass() {
        cout << "destructor!\n";
    }
    // 拷贝构造
    TestClass(const TestClass& that) :m_number(that.m_number) {
        cout << "copy constructor!\n";
    }
    // 赋值操作符
    TestClass& operator=(const TestClass& tc) {
        cout << "operator= is called\n";
        if (this == &tc)
            return *this;
        m_number = 0;
        m_number = tc.m_number;
        return *this;
    }
    int m_number;
};
TestClass tcFactory()
{
    TestClass tc(10);
    return tc;
}
int main()
{
    {
        TestClass tc = tcFactory();
    }
    return 0;
}

// 其他代码不变，只增加移动构造与移动赋值处理函数
// 移动构造
TestClass(TestClass&& rr):m_number(rr.m_number) {
    // 如果这里是指针的变量的话则可以避免指针重复释放的问题
    rr.m_number = 0;
    cout << "move constructor!\n";
}
// 移动赋值
TestClass& operator=(TestClass&& rr) {
    cout << "move operator= is called\n";
    if (this == &rr)
        return *this;
    // 此步骤相当于对源指针的释放
    m_number = 0;
    m_number = rr.m_number;
    return *this;
}

int main()
{
    vector<int> v{ 1,2,3,4 };
    // 拷贝构造
    vector<int> v1 = v;
    // 移动构造
    vector<int> v2 = std::move(v);
    cout << "v size():" << v.size() << "\n";
    cout << "v2 size():" << v2.size() << "\n";
    return 0;
}

class TestClass
{
public:
	// 这么写在 VS中也会提示 	C26478 不要对常量变量使用 std::move
    TestClass(const string& str) :m_str(std::move(str)) {
    }
    string m_str;
};
int main()
{
    string str = "sss";
    TestClass tc(str);
    cout << tc.m_str << "\n";
    // 此处应该输出空，但实际并非如此 , 两个输出都是 sss
    cout << str << "\n";
    return 0;
}