call function with cout inside a cout statement

Question

Firstly please have a look at some simple codes that my questions derived from.

#include <iostream>
#include <string>
using namespace std;

string get_something() 
{ 
    cout << "output something"; 
    return " and return something"; 
}
    
void print_something() 
{
    cout << "print something";
}

int main()
{
    cout << get_something();    // will work 
    cout << print_something();  // will NOT work

return 0;
}

The only different thing I notice between get_something() and print_something() is that one is a return type and one isn't. As you can see I have added comments indicating that which one will work and not work.

However, I am really not sure what is happening behind the scene that makes it one possible and the other not possible. I am not even sure how I should go about and search for this kind of question too.. so here I am asking a question.

Please enlighten me..

edit: I am confused that how it is possible to do cout after cout.. both of the functions do that but one of them works and the other doesn't.

Answer 1

This seems to be a very common misunderstanding among beginners. Printing something via cout is not the same as returning a value from a function. Thats completely orthogonal things.

You can write:

std::string returned_value = get_something();
std::cout << returned_value;

But you cannot write:

??? returned_value = print_something();   
std::cout << returned_value;

because print_something() does not return anything! void denotes the absence of a type. You cannot have an object of type void .

On the other hand, when you call a function, you can use the returned value (above), or you can ignore it, so this is correct code:

 print_something();       // prints something
 get_something();         // also print something and returned value is ignored

Note that the function get_something should get a better name, because it is not just "getting" a value. How about print_and_return_something() ?

PS:

What I am really confused about is that, how is it possible to do a cout after a cout? Am I just missing what cout actually does?

Not sure If I understand, but I will try... std::cout is an object of type std::ostream . It has an operator<< that you can call, similar to calling methods of other objects. The following two are identical and just use different syntax:

std::cout.operator<<( "Hello World");
std::cout << "Hello World";

When you call print_something() then first the function is executed, then the return value is returned to the caller and execution continues with the caller. This:

std::cout << get_something(); 

is more or less the same as (well, its a crude simplification, but should be ok here):

// inside get_something
std::cout << "output something"; 
// return value
std::string result{"output something"};
// now execution continues in caller
std::cout << result;

Calling cout after cout is no different from calling some other function. Suppose you have a function print() that prints something then you can write

std::string print_and_return() {
        std::string x{"Hello World"};
        print(x);
        return x;
}

The caller can do

std::string x = print_and_return(); // <- this already calls print()
print(x);                           // now we call it again

This is more or less the same as yours, just that I used some hypothetical print() instead of std::cout::operator<< .

Answer 2

Both your functions have a return type. It's just that one of them has a void return type.

The std::ostream class does not have an overload for << that takes a void type. This is sensible - what would be written to the stream in that case?

( cout is an instance of std::ostream that typically writes itself to the standard output which is normally the shell you're using to launch the program.)

Answer 3

Because print_something() has nothing to return, and cout want something to write to the console (the return value it is expecting). Therefore, it will give error.

get_something() , on the other hand, has something to return. So after executing it's rest of line (except return statement) it return the string, which gets printed by cout

Answer 4

get_something() returns something (what seems to be accepted by cout ), so cout will receive the returned thing and will work.

On the other hand, print_something() returns nothing (because its return type is void ), so cout cannot receive anything to print and won't work.

Answer 5

cout is a stream object.and we use << (insertion operator) to insert value like String,float,Int etc to it which will be displayed in output Screen.Since print_something() is not returning any value so nothing is inserted in stream,That's why it is not working.

I recommend you to read about Streams in c++..