When there are pointer variable members in the custom type, the return value and parameters of the assignment operator overload must be reference types

Problem description ： For custom types that contain Pointer type member variable When , Assignment operator overload Time function Return value and Parameters Why does the type have to be Reference type ？

#include<iostream>
using namespace std;

// Assignment operator overload 
class Person
{
    
public:
    Person(int age)
    {
    
        m_Age = new int(age);

    }

    Person & operator=(Person & p) //14 That's ok 
    {
    

        // If there is unreleased heap space , Release first 
        if (m_Age != NULL)
        {
    
            delete m_Age;//pos1
            m_Age = NULL;
        }

        m_Age = new int(*p.m_Age);// notes ： Here, the newly allocated memory space address may be the same as the above pos1 The address of the memory space released at is the same （ The reason for the compiler ）
        return *this;
    }

    ~Person()//28 That's ok 
    {
    
        if (m_Age != NULL)
        {
    
            delete m_Age;
            m_Age = NULL;
        }
    }

    int* m_Age;
};

void test01()
{
    
    Person p1(18);
    Person p2(20);
    Person p3(30);

    p3 = p2 = p1; //46 That's ok 

    cout << "p1 The age is ：" << *p1.m_Age << endl;

    cout << "p2 The age is ：" << *p2.m_Age << endl;

    cout << "p3 The age is ：" << *p3.m_Age << endl;
}

int main(void)
{
    
    test01();
    system("pause");
    return 0;
}

• case 1:

  The return value must be of reference type The reason of ： The assignment operator is overloaded in , If the return value is a value type, that is 14 That's ok Turn into

Person operator=(Person & p)

That is, the return value is of value type , Then the return value type after the function call , For the call stack of external functions, it is a Temporary objects .

​ and The temporary object is a right value （ It doesn't exist in the stack memory , The life cycle ends after a complete expression , namely p3 = p2 = p1; End after statement execution ）, Objects that do not exist in stack memory cannot be referenced with ordinary lvalues （ such as Person & ） To quote , therefore 46 That's ok Of p2 = p1 The result is a Temporary objects , That is, a Right value ,p3=(p2 = p1) Call to p3.operator(p2=p1 Returned object ),Person & Is an lvalue reference , and p2 = p1 The return result is the right value （ The life cycle is short ）, After statement end ,p3 Facing the danger of ineffectiveness , So the compiler is not allowed to pass .

​ contrary , if The return value is the reference type That is to say, what is returned is Person Object itself Not temporary objects , be p2 = p1 Back to a The left value ,p3 = (p2 = p1), Call to p3.operator(p2=p1 Returned object ) Is to give a Space that can exist for a long time in memory Alias to manipulate , So it can be compiled .

• case2： Parameter type must be reference type The reason of ： The parameter type is the value type, that is 14 That's ok Turn into

   Person & operator=(Person  p)

p3 = p2 = p1 Execute first p2 = p1, That is equivalent to p2.operator=(p1), The object passed in is p1 A copy of , be aware One of the objects is Pointer variable member m_Age, When copying Directly copy its value , meanwhile , The parameter passed in is Local objects , It will be released after the function is executed , Then it will call 28 Line destructor , Now p1 Of m_Age Memory space will be released , Followed by execution p3 = (p2 = p1), That is to call p3.operator=(p2=p1 Returned object ), Will be p2 Of m_Age Memory space is released , But when 46 Yes p3 = p2 = p1 After execution , Then the corresponding destructor will be executed again , take p1 Of m_Age To release , The memory space has been released at the end of the function call , As a result Secondary deconstruction （ Release memory space repeatedly ）.

​ But the specific implementation process , It is suggested that you should clearly understand the cause of this exception through your own debugging .