Using cout to display function return output

[x-ice]

If i have the following code:

string clientName = "Dan";

string getClientName(void)
{
     return clientName
}

How can i use cout to display this information?

p.s. I tried cout << getClientName() << endl; but it didnt work.

[twanvl]

What does "didn't work" mean? I see nothing wrong with that code, except you are missing a ";" after "return clientname".

[x-ice]

#include <iostream>
#include <string>

using namespace std;

string clientName = "Dan";

void main(void)
{
	cout << getClientName() << endl; //This is Line 10
}

string getClientName()
{
     return clientName;
}

When i use this code I receive an error on line 10, the error is 'error C3861: 'getClientName': identifier not found'

[CornedBee]

It doesn't know of the function getClientName() at the point you call it. You need a function declaration.

And main returns int, not void.

[x-ice]

It doesn't know of the function getClientName() at the point you call it. You need a function declaration.

And main returns int, not void.

According to my book (Thinking in C++) and my programming lecturer main can return void.

How do i use a function declaration?

[Harsh Gupta]

"Main returning void" were supported by older version of C++ compilers. If you are using ANSI compliant C++ compilers then Main would not return void. Atleast GCC and Borland compilers do not return void types.

Function declaration is done above main() just like you have declared the string variable clientName.

#include <iostream>
#include <string>

using namespace std;

string getClientName();
string clientName = "Dan";

void main(void)
{
	cout << getClientName() << endl; //This is Line 10
}

string getClientName()
{
     return clientName;
}

or if you define the function above main() then you would not need function declaration.

#include <iostream>
#include <string>

using namespace std;

string clientName = "Dan";

string getClientName()
{
     return clientName;
}


int main(void)
{
	cout << getClientName() << endl; //This is Line 10
	cin.get();
	return 0;
}

[x-ice]

"Main returning void" were supported by older version of C++ compilers. If you are using ANSI compliant C++ compilers then Main would not return void. Atleast GCC and Borland compilers do not return void types.

Function declaration is done above main() just like you have declared the string variable clientName.

#include <iostream>
#include <string>

using namespace std;

string getClientName();
string clientName = "Dan";

void main(void)
{
	cout << getClientName() << endl; //This is Line 10
}

string getClientName()
{
     return clientName;
}

or if you define the function above main() then you would not need function declaration.

#include <iostream>
#include <string>

using namespace std;

string clientName = "Dan";

string getClientName()
{
     return clientName;
}


int main(void)
{
	cout << getClientName() << endl; //This is Line 10
	cin.get();
	return 0;
}

I am using Visual Studio 2005, so whatever compiler comes with VC++ 2005.

[Harsh Gupta]

Sorry, I am not sure about MS based C++ compilers.

So, is it still not working??

[x-ice]

Sorry, I am not sure about MS based C++ compilers.

So, is it still not working??

I have made the changes you said about, and its works. Thanks for your help.

[CornedBee]

According to my book (Thinking in C++) and my programming lecturer main can return void.

Throw away both your book and your lecturer.

http://faq.cprogramming.com/cgi-bin/...&id=1043284376
http://users.aber.ac.uk/auj/voidmain.shtml

[x-ice]

Throw away both your book and your lecturer.

http://faq.cprogramming.com/cgi-bin/...&id=1043284376
http://users.aber.ac.uk/auj/voidmain.shtml

Ok i get your point, i will use int main() instead