Is in operator overloading the left side of the operator is implicitly passed to function?

Question

I Have some doubt in operator overloading in c++.

#include<iostream>

using namespace std;


class ex
{
    int i;
public:
    ex(){}
    void operator+(ex ob);
    void seti(int x);
    int geti();
};

void ex::operator+(ex ob)
{
    this->i+=ob.i;
}

void ex::seti(int x)
{
    i=x;
}

int ex::geti()
{
    return i;
}

int main()
{
    ex ob1,ob2;

    ob1.seti(2);
    ob2.seti(2);

    ob1+ob2;

    cout<<"ob1:"<<ob1.geti();
    cout<<"\nob2:"<<ob2.geti();

    return 0;
}

output:

ob1:4
ob2:2
Process returned 0 (0x0)   execution time : 0.400 s
Press any key to continue.

Here the left of the operator + that is ob1 is passed implicitly (this pointer) to the function and ob2 is passed through the parameter to the function I am OK with that.

But see the following code.

#include<iostream>

using namespace std;


class ex
{
    int i;
public:
    ex(){}
    void operator++();
    void seti(int x);
    int geti();
};

void ex::operator++()
{
    this->i++;
}

void ex::seti(int x)
{
    i=x;
}

int ex::geti()
{
    return i;
}

int main()
{
    ex ob1;

    ob1.seti(2);

    ob1++;

    cout<<"ob1:"<<ob1.geti();

    return 0;
}

When i compile above Shows This error:

D:\Users\srilakshmikanthan.p\Documents\source code\ex.cpp||In function 'int main()':|
D:\Users\srilakshmikanthan.p\Documents\source code\ex.cpp|37|error: no 'operator++(int)' declared 
for postfix '++' [-fpermissive]|
||=== Build failed: 1 error(s), 0 warning(s) (0 minute(s), 2 second(s)) ===|

When i change the line ob1++; to ++ob1; It will compile successfully and shows result 3.

My doubt is In operator overloading the left side of the overloaded operator is passed through implicitly(this pointer) to the function but why this is not happens here. The compiler says change the line ob1++; to postfix like ++ob1; but the object in right.

Please help me.

Does this answer your question? [C++: overloading ++ for both pre and post increment](https://stackoverflow.com/questions/15244094/c-overloading-for-both-pre-and-post-increment) — Renat, Dec 07 '19 at 10:57
Does this answer your question? [What are the basic rules and idioms for operator overloading?](https://stackoverflow.com/questions/4421706/what-are-the-basic-rules-and-idioms-for-operator-overloading) — Richard Critten, Dec 07 '19 at 10:57

score 0 · Answer 1 · answered Dec 07 '19 at 11:03

This member function

void operator++();

defines a pre-increment operator ++ that should be declared and defined like

ex & operator++();

and

ex & ex::operator++()
{
    this->++i;

    return *this;
}

The declaration and the definition of the post-increment operator can look like

ex operator ++( int );

and

ex operator ++( int )
{
    ex obj = *this;
    i++;

    return obj;
}

So the compiler issues an error because in the program there is used the post-increment operator that was not declared and defined for the class ex.

When an operator is a member function then the left operand of the operator is implicit argument *this of the function

Is in operator overloading the left side of the operator is implicitly passed to function?

1 Answers1