CODEWITHSUNDEEP
c++referenceoperator-overloading
L-caesar
L-caesar

Reputation: 75

what is the differences between returning reference or not in operator overloading C++

I've tried to figure out what is the purpose of & on the return type. I mean,consider the code below, what happens if i delete & from the operator overloading function.

    class Container
{


    public:
        int numElems;
        int *data;


    Container(int n):numElems(n){data=new int [numElems];}
    Container & operator=(const Container &rhs)
    {
        if(this!=&rhs)
        {
            if(data!=NULL)
                delete  [] data;
        numElems=rhs.numElems;
        data=new int [numElems];
        for (int i=0;i<numElems;i++)
        {   
            data[i]=rhs.data[i];    
        }
            return *this;
        }
    }

};

I deleted it and compile it ,it compiled without any errors.Actualy it gives the same result in both cases for an example main:

int main()
{
Container a(3);
Container b(5);
Container c(1);
cout<<a.numElems<<endl;
cout<<b.numElems<<endl;
cout<<c.numElems<<endl;
a=b=c;
cout<<a.numElems<<endl;
cout<<b.numElems<<endl;
cout<<c.numElems<<endl;
return 0;
}

So, is there anyone who can help me about the purpose of & on the left side ? Thanks in advance.

Upvotes: 1

Views: 113

Answers (3)

Marcus S.
Marcus S.

Reputation: 543

class foo {

    public:

        int val;

        foo() { }
        foo(int val) : val(val) { }

        foo& operator=(const foo &rhs) {
            val = rhs.val;
            return *this;
        }

        foo& operator++() {
            val++;
            return *this;
        }

};


void main() {
    foo f1(10), f2;
    (f2 = f1)++;
    std::cout << f1.val << " " << f2.val << std::endl;
}

Output:

10 11

Output when removing reference:

10 10

Upvotes: 1

Marc Claesen
Marc Claesen

Reputation: 17026

If you don't return a reference, you implicitly make an extra unnecessary copy.

Upvotes: 0

aaronman
aaronman

Reputation: 18750

Returning a reference is much faster than returning a value for a large object. This is because under the hood a reference is just a memory address whereas if you return it by value it requires a deep copy

Upvotes: 0

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