friend declaration in C++
- by Happy Mittal
In Thinking in C++ by Bruce eckel, there is an example given regarding friend functions as
// Declaration (incomplete type specification):
struct X;
struct Y {
void f(X*);
};
struct X { // Definition
private:
int i;
public:
friend void Y::f(X*); // Struct member friend
};
void Y::f(X* x) {
x->i = 47;
}
Now he explained this:
Notice that Y::f(X*) takes the address of an X
object. This is critical because the compiler always knows how to
pass an address, which is of a fixed size regardless of the object
being passed, even if it doesn’t have full information about the size
of the type. If you try to pass the whole object, however, the
compiler must see the entire structure definition of X, to know the
size and how to pass it, before it allows you to declare a function
such as Y::g(X).
But when I tried
void f(X);
as declaration in struct Y, it shows no error.
Please explain why?
