C++ - ostream (<<) overloading

Question

I was wondering if there is any way to overload the << operator for a class without declaring it as a friend function. My professor said this is the only way to do it, but I wanted to know if there is another way that he was unaware of.

http://stackoverflow.com/questions/4421706/operator-overloading/4421719#4421719 Input and Output Operators :) — lcjury, Apr 03 '14 at 01:48

score 2 · Answer 1 · answered Apr 03 '14 at 01:49

2

There is no need to make the operator<< function a friend of the class as long as everything you want to be output is accessible through the public interface of the class.

answered Apr 03 '14 at 01:49

R Sahu

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score 1 · Accepted Answer · answered Apr 03 '14 at 01:57

You need to declare it is as friend function if and only if you need access to it's private members.
You can always do this without using friend function if:
1) No private member access is required.
2) You provide a mechanism to access your private member otherwise. e.g.

class foo
{
    int myValue;
    public:
    int getValue()
    {
        return myValue;
    }
}

score 0 · Answer 3 · answered Apr 03 '14 at 01:43

Yes, you can

std::ostream& operator<<(std::ostream &stream, WHATEVER_TYPE var) {
    std::string str = somehowstringify(var);
    return stream << str;
}

Note however that by virtue of it being a non-member non-friend function it can of course only access the public interface of std::ostream, this usually isn't a problem.

score 0 · Answer 4 · answered Apr 03 '14 at 01:50

0

Yes, one way to do it is like this:

class Node
{
public:
    // other parts of the class here
    std::ostream& put(std::ostream& out) const { return out << n; };
private:
   int n;
};

std::ostream& operator<<(std::ostream& out, const Node& node) {
    return node.put(out);
}

answered Apr 03 '14 at 01:50

Edward

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The member one is very confusing, you're saying that `ostream << Node` is the same as `Node << ostream`, the first doesn't make sense to me. – Cramer Apr 03 '14 at 02:36
No, the member function is `put` and has a return type of `std::ostream&`. The `operator<<` uses the `put` member function so that it does not have to expose anything about the internals of `Node`. You'll see that `n` is only used within the body of the member function and is not visible to the other function. – Edward Apr 03 '14 at 02:41

score 0 · Answer 5 · answered Apr 03 '14 at 02:01

As R Sahu has pointed out, the requirement is that the operator should be able to access everything it has to display.

Here are a few possible options

1.Adding the overloaded function as a friend function

2.Making all the required data members of the class accessible for the function using either public accessor methods or public data members

class MyClass {
   private:
   int a;
   int b;
   int c;
   public:
   MyClass(int x,int y,int z):a(x),b(y),c(z) {};
   MyClass():a(0),b(0),c(0) {};
   int geta() { return a; }
   int getb() { return b; }
   int getc() { return c; }
};

std::ostream& operator<<(std::ostream &ostr,MyClass &myclass) {
   ostr << myclass.geta()<<" - " << myclass.getb() << " - " << myclass.getc() ;
   return ostr;
}


int main (int argc, char const* argv[])
{
   MyClass A(4,5,6);
   cout << A <<endl;

        return 0;
}

3.Add a public helper function , say output with the signature std::ostream& output(std::ostream& str) and use it later in the operator function.

C++ - ostream (<<) overloading

5 Answers5