Overloading assignment operator: different datatypes - impossible?

Question

I am trying to make something that is probably impossible to do with operator overloading. I am particularly interested in overloading assignment operator, which would receive another data type value as its right hand value. It should look like this:

MyClass myclass = "hello world";    <--- Wrong?

MyClass myclass2;
Myclass myclass = myclass2;         <--- Right?

Then MyClass object should receive the string and handle it. Unfortunately, from what I read it is only possible to assign the same data type value to a custom made class. Is that true or was I mistaken?

This is the code I currently have:

class MyClass {
public:
    MyClass() {};
    virtual ~MyClass();
    MyClass& operator = (const MyClass&);
private:
    char* string;
};

MyClass& MyClass::operator= (const MyClass& inc){
    string = inc;
}

int main(int argc, char** argv) {
    MyClass myclass = "hello world";
    std::cout << myclass;
}

As you can see, I want also to cout the object as a string. Basically, I want my custom class to be treated as a string. My search in Google and StackOverflow search engines denied my wish, but is it actually like that or is there a workaround?

Looking forward to hear from you and thank you very much for help in advance!

EDIT: Rollie fixed the main problem. However, how would we cout the string value of the custom object MyClass? Is that even possible, since object output is just the memory address of the object being outputted?

you were mistaken, you can have any type on the right hand side — Mooing Duck, Jul 25 '13 at 19:45

Rollie · Accepted Answer · 2013-07-25T19:50:45.913

3

MyClass myclass = "hello world"; is not an assignment - it's a constructor, and is logically equivilent to MyClass myclass("hello world");. Overload both the constructor and assignment operators, and you'll have the behavior you are looking for!

Also some comments:

1) It isn't a great idea to name your member variable string - this is the name of a very common STL type

2) setting this variable directly is a time bomb; you aren't copying the value of the string, you are just copying the pointer. switch to std::string so the copies happen automatically

3) to get cout to work properly, see the answer at stackoverflow.com/questions/5508857/how-does-cout-actually-work

edited Jul 25 '13 at 19:50

answered Jul 25 '13 at 19:02

Rollie

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Or just overload the constructor and let the conversion happen for the assignment operator. – chris Jul 25 '13 at 19:05
Good point Chris, definitely the easier way to go, but less optimal due to the extra object construction I believe. – Rollie Jul 25 '13 at 19:07
silly me :) thanks. worked too much in other programming languages – Jonas Hoffmann Jul 25 '13 at 19:07
@Rollie, Depends on just what kind of difference it actually makes. I wouldn't go to the extra trouble unless I knew it was worth it. – chris Jul 25 '13 at 19:09
By the way, how we would make it to `cout` the string value of `MyClass` object? – Jonas Hoffmann Jul 25 '13 at 19:11
Is that even possible? When we usually `cout` an object, it outputs the memory location of the object. – Jonas Hoffmann Jul 25 '13 at 19:18
see the answer at http://stackoverflow.com/questions/5508857/how-does-cout-actually-work – Rollie Jul 25 '13 at 19:26
1

@Rollie: Making the class castable to a `const char*` probably isn't a good idea. A better idea is to overload `operator<<` – Mooing Duck Jul 25 '13 at 19:47

Overloading assignment operator: different datatypes - impossible?

1 Answers1