C++ void type - how?

Question

typedef void(Object Sender) TNotifyEvent;

This is what I'm trying to do from Delphi to C++, but it fails to compile with 'type int unexpected'.

The result should be something I can use like that:

void abcd(Object Sender){
  //some code
}

void main{
   TNotifyEvent ne = abcd;
}

How do I make such type(of type void)? I'm not very familiar with C++.

Answer 1

If what you want is to define the type of a function that takes an Object as parameter and returns nothing, the syntax would be:

typedef void TNotifyEvent( Object Sender );

EDIT, as answer to the comment.

Yes, you can define the type of a function, and that type can later be used in different contexts with different meanings:

TNotifyEvent func;          // function declaration (weird as it might look)
                            // same as: void func( Object Sender );
TNotifyEvent *fp = func;    // function pointer declaration -- initialized with &func
void func( Object Sender )  // cannot use the type when defining the function
{}
void foo( TNotifyEvent f ); // compiler translates to TNotifyEvent * f
                            // just as 'int a[5]' is converted to 'int *a' 
                            // in function parameter lists.

Answer 2

Do you want a pointer to a function which takes an Object and returns nothing?

 typedef void (*TNotifyEvent)(Object Sender);

Answer 3

No, because there is no common "object" class in C++. What you probably want there is a callback (Looking at events...), for which you need a functor (a class overloading operator()) or a function pointer.

Here's an example using the function pointer:

typedef void (*TNotifyEvent)(void *sender);