Why does overloading ostream's operator<< need a reference "&"?

Question

I've been learning C++.

From this page, I understood that overloading "<<" operator of ostream can be done in this way.

ostream& operator<<(ostream& out, Objects& obj) {
    //
    return out;
} 
//Implementation

and

friend ostream& operator<<(ostream& out, Object& obj);
//In the corresponding header file

My question is... why does this function need "&" at the end of ostream and Object?

At least I know that "&" is used to...

Take the address of a value
Declare a reference to a type

However, I think neither of them applies to the overloading described above. I've spent a lot of time in googling and reading a text book, but I can't find the answer.

Any advice will be appreciated.

Passing the object by reference is to prevent copying it while ensuring the object reference is always valid (comparing to pointer). — Alexander V, May 16 '15 at 05:27

Martin York · Accepted Answer · 2015-05-16T05:47:00.523

why does this function need "&" at the end of ostream and Object?

Because you are passing them by reference.
Why are you passing them by reference. To prevent a copy.

ostream& operator<<(ostream& out, Objects const& obj)
                             //           ^^^^^       note the const
                             //                       we don't need to modify
                             //                       the obj while printing.

The obj can be copied (potentially). But what if it is expensive to copy. So best to pass it by reference to prevent an unnecessary copy.

The out is of type std::ostream. This can not be copied (the copy constructor is disabled). So you need to pass by reference.

I normally declare the stream operator direcctly in the class declaration:

class X
{
    std::string    name;
    int            age;

    void swap(X& other) noexcept
    {
        std::swap(name, other.name);
        std::swap(age,  other.age);
    }
    friend std::ostream& operator<<(std::ostream& str, X const& data)
    {
        return str << data.name << "\n" << age << "\n";
    }
    friend std::istream& operator>>(std::istream& str, X& data)
    {
        X alt;
        // Read into a temporary incase the read fails.
        // This makes sure the original is unchanged on a fail
        if (std::getline(str, alt.name) && str >> alt.age)
        {
            // The read worked.
            // Get rid of the trailing new line.
            // Then swap the alt into the real object.
            std::string ignore;
            std::getline(str, ignore);
            data.swap(alt);
        }
        return str;
    }
};

More to the point, you wouldn't _want_ the stream to be copied even if doing so made any technical sense. — Lightness Races in Orbit, Sep 05 '16 at 19:54

score 0 · Answer 2 · answered May 16 '15 at 05:35

Example

void f(int& index)
{
    index = 3;
}

means f is a function with an int parameter that is passed by reference. So after

int a = 4;
f(a);

a has a value of 3. For the operator you mentioned this means that the ostream as well as Object may be changed during the execution of the operator (as some kind of function).

Why does overloading ostream's operator<< need a reference "&"?

2 Answers2