'<<' operator

If someone can help me I appreciate it. The following code give me an error. I could not figure it out

Code:

---

#include <iostream>
using namespace std;

void A_print(){
    cout << "A_print" << endl;
}

int B_print(int x){
    int y = x/2;
   
    return y;
}
int main(){
   
    A_print();
    cout << "2nd  " << A_print() << endl; // ?? right this line if I change A_print() to B_print(6) it is ok

    cout << endl;
    system("pause");
    return 0;
}

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Thank you

A_print's return type is void, so you cannot print its return value as it has none.

One solution is to call A_print separately.

Yeah it's a good idea to post the error:

Code:

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In function 'int main()':
Line 16: error: no match for 'operator<<' in 'std::operator<< [with _Traits = std::char_traits<char>](((std::basic_ostream<char, std::char_traits<char> >&)(& std::cout)), ((const char*)"2nd  ")) << A_print()'
compilation terminated due to -Wfatal-errors.

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If we ignore that you called a function for a moment we can boil your code down to this

Code:

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cout << "2nd " << cout << "A_print" << endl << endl;

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cout can't exactly print itself; and you don't return a value from A_print() to display anyway. There is a way to make this syntactically correct. Break it up.

Code:

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cout << "2nd ";
A_print();
cout << endl;

---

Thank you guys,

Actually I figure it out this:

Functions like int, float, functions does return something, has to be called with "cout".
Functions like void, does not return something has to be called without "cout".

Am I right?

You're right.

If the function returns a value you're free to do whatever you please with it, as if it was a variable.
If it's void you mustn't include cout before you call it. It does it itself.

Thanks, that help a lot and make sense for me now.

Appreciate your time guys.
:)

Quote:

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Originally Posted by valentin

Thank you guys,

Actually I figure it out this:

Functions like int, float, functions does return something, has to be called with "cout".
Functions like void, does not return something has to be called without "cout".

Am I right?

---

Well, to be more precise, you can include a call to a function that returns something other than void in a cout statement. It doesn't mean you have to.
You can't, however, include a function call to a function returning void in a cout statement. The reason is that you are essentially telling the compiler to pass whatever the function is returning to cout, but if it returns nothing, how can it do that?