Easy Peasy Function Problem

Just starting to look at functions and i wanted to write one that would go on to a new line. I know this code is really impractical but if you know what i am trying to do would you be a pal and show me the way. The less you change my code the more help it will be to me.

// This is my first function (no ........ sherlock)

#include <iostream.h>

void nline(void); // function prototype

int main()
{



cout << "Hello"
<< nline(void)
<< "Kiddies"
<< nline(void)

return 0;
}

void nline(void) // newline function
{
cout <<endl;
return ;
}

Any other constructive help about functions would be appreciated.
(an url maybe)

Thanks

You don't need the << before the nline() function -

cout << "Hello" ;
nline();
cout << "Kiddies";
nline();


You should either use '\n' or endl, it would make your life easier.

Zen can you somehow return on ostream object though?

Would you have to overload the '<<' operator?

Zen you are a star!

I know the program was of sod all use but it was the easiest function i could think to write so the fact that i could'nt do that was some what disturbing.

Aww, come on, no answer? Oh well.

Yes but you can only overload the << operators for user defined types, not primitives or functions. It works like this -

Code:

---

#include <iostream>

using namespace std;

class t
{
public:
    int i;
};

ostream& operator<< (ostream& os,t test)
{
    return os<<test.i;
}


int main()
{
   
   
    t test;
    test.i=10;

    cout << test << endl;

    return 0;
}


---

If you wanted to return an ostream just to create a newline , I think you'd have to still call a function -

Code:

---

#include <iostream>

using namespace std;


ostream& nline(ostream& os)
{
    return os<<"\n";
}
int main()
{
   
    nline(cout);

    return 0;
}


---

So you can't return a manipulator in a compound cout statement? Thanks, that code helps.

If by compound statement you mean chaining, then you can do that. That's why the ostream is returned rather than just outputing the type in the function.

Code:

---

#include <iostream>

using namespace std;

class t
{
public:
    int i;
};

ostream& operator<< (ostream& os,t test)
{
    return os<<test.i;
}


int main()
{
   
   
    t test;
    test.i=10;

    t test2;
    test2.i=20;

    cout << test << endl <<test2<< endl;

    return 0;
}


---

Yeah I can see how a copy of the class data is returned and I guess the manipulator '\n' should just be called with 'endl' rather than trying to return it with a function that is part of a 'cout' chain.

This was confusing at first though:

Code:

---

ostream& operator<< (ostream& os,t test)
{
    return os<<test.i;
}

---

Because I would have thought that 'ostream& os' was not a parameter since 'ostream&' was a return value. But than again i'm not used to working with C++ streams.

There's nothing stoppping you returning an endl or '\n' included in the ostream but when overloading operators it's best to imitate the default behavior as when you pass an int into an ostream -

int i=0;
cout << i;

it doesn't get an endl or '\n' tagged onto the end of it.

It's up to the user(programmer) to decide if they want to put one on and you may have other functions that rely on there not being a newline.

Code:

---

ostream& operator<< (ostream& os,t test)
{
    return os<<test.i << endl;
}

---

Actually this does work. Yes true ofcourse there is no need to do this, that wasn't what I was after. Anyway this is good.