C ++ iostream运算符覆盖了函数的返回类型

Question

I'm still in learning phase of basic formats and commands of C++. I'm now at class operator function overloading and came to << and >> . My question is: when they are defined in friend functions such as below:

ostream &operator << ( ostream &output, const PhoneNumber &number )

and are called with PhoneNumber class phone like this:

cout << phone << endl;

Why is the friend function returning ostream& ? I mean when a function returns a value of a particular type, it is generally received by a fundamental type variable such as bool, int, char, string, and etc. However, for ostream and istream , the returned type of ostream& is not being saved. Then, in this case, shouldn't it be void (carry out the task and terminate without returning any values)?

Answer 1

Because otherwise you would be able to chain the calls to operator<< . This:

cout << phone << endl;

is parsed as:

(cout << phone) << endl;

and resolves as:

operator<<(cout, phone).operator<<(endl);

So it first calls operator<<(cout, phone) , which returns cout , which then allows the second << to call cout.operator<<(endl) .

If operator<< returned eg void , the second << would try to call operator<<(void, endl) which would not compile.

Answer 2

"Why is the friend function returning ostream& ?"

To make this part of the call chain working

phone << endl;

The ostream& reference is passed through all of these function calls, and thus the operator<<() function can be called again on the result.

Answer 3

Why is the friend function returning ostream&?

To allow operators to be chained together. The operators return a reference to the same stream that was passed in. Without that return reference, you would not be able to call << endl on the return value of cout << phone .

I mean when a function returns a value of a particular type, it is generally received by a fundamental type variable such as bool, int, char, string, and etc.

That is optional. A return value is temporary and goes out of scope at the end of the statement, if it is not assigned to a variable to extend its lifetime.

However, for ostream and istream, the returned type of ostream& is not being saved.

It does not have to be. It just has to stay alive long enough for the next operator to be called on it, if the final ; has not been reached yet.

Then, in this case, shouldn't it be void (carry out the task and terminate without returning any values)?

No.