operator -overloading in C++

Question

I have a assignment that construct Operator overloading with bitwise.I try to construct operator<< to shift integer left,but when I declare function is: friend const int operator<<(const int& , int& ); the complier notice error nomember operator requier a parameter with class or enum type.. So I don't operator overloading with operator<<. This is my code:

#pragma once//bitwise.h
#include<iostream>
using namespace std;
class bitwise
{   private: 
    int n;
public:
    bitwise(int n){
        this->n = n;
    }
    friend const int operator<<(const int& , int& );//complier notice error here.
};
const int operator<<(const int& n, int& x){
    int temp=n*pow(2, x);
    return temp;
}

Yes it's the rule. You can only overload operator with user-defined type. — songyuanyao, Apr 10 '16 at 14:38

score 0 · Accepted Answer · answered Apr 10 '16 at 14:40

0

The way you are creating the operator, it is trying to redefine what it means to use shift-left with two integers, and not with your class. You want something like this:

friend int operator<<(const bitwise& , int);

answered Apr 10 '16 at 14:40

Vaughn Cato

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thanks, I have just done. But I read in my document "operator<<" can declared as: "friend T operator<<(const T& , T&);". Is this wrong? – Nguyen Van Binh Apr 10 '16 at 14:57
@NguyenVanBinh: That's true as long as `T` is not a built-in type. – Vaughn Cato Apr 10 '16 at 15:01
Ok, I understand. thank you so much. – Nguyen Van Binh Apr 10 '16 at 15:07

operator -overloading in C++

1 Answers1