Question

I’ve been working on this school assignment. The assignment told us to make an object which had it’s output operator ( << ) overloaded.
Here’s my code:

#include <ostream>
using namespace std;

template <class T>
class CustomObject {

        string print() {
            string text = "";
            for (int i = 0; i < num_items(); i++) {
                text += queue[i];
                text += " | \n";
            }
            return text;
        }

        friend std::ostream& operator <<(std::ostream &output, CustomObject &q) {
            output << "" << q.print();
            return output;
        }
}

So I instantiate this object like this:

CustomObject<int> co();

and call its output method:

std::cout << co();

Which would inevitably call the print method, and return the string to the default output stream.

But, there’s no visible output in my console/debugger.

What am I missing here?

PS this is not the complete class, it’s generic because of several other methods and functionality that is not necessary to be shown here.

PPS the num_items() and queue variables are part of said rest, this class is a PriorityQueue object. So, queue is an array of the specified type (hence the generic declaration) and num_items() just returns the count of the array.

Answer 1

CustomObject<int> co();

That’s a function declaration. Leave out the parenthesis.

std::cout << co();

Why are you appling operator() to co? Again, leave out the parenthesis. This should work:

CustomObject<int> co;
std::cout << co;

Alas, building and returning a string from a print method is hardly idiomatic C++. Here is what I would do:

template <typename T>
class CustomObject
{
    // ...

public:

    void print(std::ostream& os) const
    {
        for (int i = 0; i != num_items(); ++i)
        {
            os << queue[i] << " | \n";
        }
    }
};

std::ostream& operator<<(std::ostream& os, const CustomObject& object)
{
    object.print(os);
    return os;
}