What is the purpose of returning an IO object in C++?

Question

I am attempting to teach myself C++ by reading through a textbook and doing practice problems and stuff, and the topic I am currently learning is a little confusing to me and I am hoping to get some clarification. I have looked online for a clear answer to my question, but have not yet found anything.

I am currently learning the details of IO Classes in the standard library, and the section I am on right now gives some examples that has functions that pass and return IO objects.

For example:

istream &get_value(istream &input)
{
    int value;
    input >> value;

    return input;
}

int main()
{
    get_value(cin);
    return 0;
}

I understand on a high-level view what is happening here. The get_value function has a reference to an input object type and it also takes in a reference to an input object, which in my example I used to commonly used cin object. I get that this function is reading input from the user in the console and is storing that input as value.

What I do not understand is what the reason for returning the input object is. Why shouldn't this function have a type void? What could the input object I am using be used for? I know I am not using it for anything right now, but what could it be used for?

Answer 1

The return value is so you can "chain" the calls between the stream operators << and >>. Operator overloading is a good motivation for this "chaining".

using namespace std;
class book {
    string title;
    string author;
public:
    book(string t, string a) : title(t), author(a) { }
    friend ostream &operator<<(ostream &os, const book &x);
}

ostream &operator<<(ostream &os, const book &x)
{
    os << x.title << " by " << x.author << "\n";
    return os;
}

int main()
{
    book b1 { "Around the World in 80 Days", "Jules Verne" };
    book b2 { "The C Programming Language", "Dennis Ritchie" };

    cout << b1 << b2; // chaining of operator<<
}

If operator<< didn't return an ostream, we would not be able to pass the modified ostream from the first operator<< to the second one. Instead we would have to write

cout << b1;
cout << b2;

The same applies for input operations, like in your case, with >>