Operator overloading assignment operator

Question

#include<iostream> using namespace std; class shared_ptr { public: int *pointer; public: shared_ptr() { pointer = new int; } ~shared_ptr() { delete pointer; } int operator* (); int* operator= (shared_ptr&); }; int shared_ptr:: operator* () { return *(this->pointer); } int* shared_ptr:: operator= (shared_ptr& temp) { return (temp.pointer); } int main() { shared_ptr s1; *(s1.pointer) = 10; cout << *s1 << endl; int *k; k = s1; //error cout << *k << endl; }

I am trying to create something like smart pointer.

I am getting the following error while trying to overload operator = .

prog.cpp:39:9: error: cannot convert ‘shared_ptr’ to ‘int*’ in assignment for the k = s1 assignment line. What am I missing here?

Can I have implementation where I dont need cast? I mean using this like pointers — Anup Buchke
– Anup Buchke, Commented Jul 4, 2013 at 21:22
to allocate memory then copy will be nice idea I think, otherwise you may victim of segfault — Grijesh Chauhan
– Grijesh Chauhan, Commented Jul 4, 2013 at 21:23

RiaD · Accepted Answer · 2013-07-04 21:23:11Z

You did provide operator = for

shared_ptr = shared_ptr

case(very strange operator btw). But you are trying to use

int* = shared_ptr

You need either getter or cast-operator in shared_ptr to make it possible

Actually you may use it like

shared_ptr s1, s2; ... int* k = (s1 = s2);

Demo

But it's absolutely ugly

Kostia · Accepted Answer · 2013-07-04 21:33:39Z

Your Operator = returns int* but you don't have a constructor that gets int*, add:

shared_ptr(int *other) { pointer = new int(*other); }

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