Arithmatic operator overloading in c++

Question

#include<iostream>
using namespace std;
class money
{
    int rs;  
    int p; 

    public:

    void setdata (int x , int y) 

     {rs=x; p=y;}

    void show() 

   { cout  <<rs  <<"."  <<p; }  

    money operator += (int a)  {

    money temp;
    temp.rs=rs+a.rs;
    temp.p=p+a.p;   
    return (temp);  
    }
};



  int main() {

    money c1,c2;

    c1.setdata(8,2);

    c2=c1.operator+=(4);

    c2.show();

}

Can someone tell me why the operator += overloading doesn't work?

My desiring output is 12.2 but the output i got is 16.2 .

I am sending 4 as argument and i want this argument is added in r (ruppee) part

`+=` is supposed to modify `this` and return a reference to `this` — 463035818_is_not_a_number, Sep 19 '18 at 10:03
Possible duplicate of [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) — 463035818_is_not_a_number, Sep 19 '18 at 10:03
please make sure the code matches the problem you describe. See also [mcve] — 463035818_is_not_a_number, Sep 19 '18 at 10:06
@user463035818 i just want to return temp into the c2 object — , Sep 19 '18 at 10:08
`temp.p=p;` you are not adding anything here. And if you fix that your operator is still doing the wrong thing. What you implemented is an `operator+` — 463035818_is_not_a_number, Sep 19 '18 at 10:08
@user463035818 sir this is my assignment questions and its always throwing wrong output — , Sep 19 '18 at 10:11
@user463035818 13.5 + 1.5 = 15.00 what just i want to show sir — , Sep 19 '18 at 10:12
i still dont get it, but probably this ` temp.rs=rs+rs;` is not what you want but rather `temp.rs = rs + a.rs;`. Currently you dont use the `a` parameter at all — 463035818_is_not_a_number, Sep 19 '18 at 10:13
@user463035818 i have just edited my code.. and it still throws a error about temp.rs = rs + a.rs; — , Sep 19 '18 at 10:15
why?? now it wont compile anymore. My mistake was to overlook that the parameter to the operator is an `int` not a `money`. If you want to add that `int` to your money then you need to add it... — 463035818_is_not_a_number, Sep 19 '18 at 10:17
`rs` is an `int`, `a` is an `int`. How do you add two `int`s? — 463035818_is_not_a_number, Sep 19 '18 at 10:19
@user463035818 affter adding it show 20.2 as an ouput it stills wrong. — , Sep 19 '18 at 10:20
hum this question is a train wreck. Here is the correct implementation: `money& operator+=(int a) { rs += a; return *this; }` and you use it as `c1 += 4;` — 463035818_is_not_a_number, Sep 19 '18 at 10:22

score 1 · Answer 1 · answered Sep 19 '18 at 10:36

1

#include<iostream>
using namespace std;
class money
{
    int rs;  
    int p; 

    public:

    void setdata (int x , int y) 

     {rs=x; p=y;}

    void show() 

   { cout  <<rs  <<"."  <<p; }  

   money& operator+=(int a) 
   { rs += a; return *this; }
};



  int main() {

    money c1,c2;

    c1.setdata(4,2);

     c2=c1+=(4);               //c2=c1.operator+=(4);

    c2.show();

}

answered Sep 19 '18 at 10:36

`c2=c1+=(4);` is correct, but it completely misses the point of having a `operator+=` instead of a `operator+`. `operator+=` modifies the left hand side and its return can be assigned to `c2` but typically you dont. Just use `c1+=4; c1.show();` – 463035818_is_not_a_number Sep 19 '18 at 11:12

Arithmatic operator overloading in c++

1 Answers1