invoking member functions without creating an object or pointer of the class?

[Attila Feher]

The whole point of using member functions is that they are invoked with
an object. If you don't want that, don't use a (non-static) member
function. Use a standalone function or a static member function.

But if that's for debugging purpose, maybe you can use some pointer
trick. But I'm not sure how valid the program would go;

test* t = 0;
t->fun();

Nope. That is illegal. Don't ever do it!

WW aka Attila

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[Arvind Sharma]

You can declare the member function as a 'static' function, then you don't
need the object to be instantiated.

In your example - if fun() is declared as 'static void fun()', then you can
call it as

test.fun()

Arvind

Is there any possibility of invoking the member functions of a class
without creating an object (or even a pointer to ) of that class.

eg.
#include <iostream.h>
class test
{
public:
void fun()
{
cout<<"Inside test::fun\n";
}
};

I want to call fun() of class test without creating
test t or even test * ptr;?
I searched on the net for the convincing answer but didn't get any.
(http://www.devx.com/tips/Tip/15846 )

Or is this a totally insane question? (I won't mind if anyone agrees
with that ..)

Thanks and Regards,
Yogesh Joshi

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[Marcus Kwok]

You can declare the member function as a 'static' function, then you don't
need the object to be instantiated.

In your example - if fun() is declared as 'static void fun()', then you can
call it as

test.fun()

You mean:

test::fun()