Question

I think the following code can be used to create manipulators.

#include<iostream> 
ostream & symbol(ostream & output)
{
  return output << "\tRs";
}

it is working fine. The following statement

  cout << "Total amount: " << 567 << symbol; 

gives the output

 Total amount: 567   Rs

But I didn’t understand why it is working.
I have the following information about operator overloading in C++.

1. only existing operators can be overloaded. New operators cannot be created.
   But the symbol is not existing operator.

2. In the statement (cout << “Total amount: ” << 567 << symbol;), it seems that << is the overloaded operator and symbol is a variable/object.
   But I didn’t declare symbol as variable/object.

3. why are they using the return statement (return output << “\tRs”;)?. I think (return “\tRs”;) or (output << “\tRs”;) should work.( I tried but not working 🙂 )

Actually I don’t know how the above code is working. Is there anybody to explain the working of the above operator overloading?

Answer 1

You are passing the function symbol to the operator <<. The << will call that function on the current ostream (with the ostream object as parameter), thus achieving the result you see. (The exact version of << called is: ostream::operator<< (ostream& ( *pf )(ostream&)); — see the reference for more info)

The return type is ostream, to allow chaining of multiple <<‘s. You would not need it technically in your particular case as << has access to the stream, but this is to keep it consistent with the operators (I think). Of course << requires this return parameter, so you have no choice 🙂