struct B
{
void (B::*pf)(int, int); // data member
B () : pf(&B::foo) {}
void foo (int i, int j) { cout<<"foo(int, int)\n"; } // target method
};
int main ()
{
B obj;
// how to call foo() using obj.pf ?
}
In above test code, pf
is a data member of B
. What's the grammar rule to invoke it ? It should be straight forward, but I am not getting a proper match. e.g. If I try obj.*pf(0,0);
then I get:
error: must use ‘.*’ or ‘->*’ to call pointer-to-member function in ‘pf (...)’, e.g. ‘(... ->* pf) (...)’
Using a pointer-to-member-function to call a function Calling the member function on an object using a pointer-to-member-function result = (object. *pointer_name)(arguments); or calling with a pointer to the object result = (object_ptr->*pointer_name)(arguments);
We can define pointer of class type, which can be used to point to class objects. Here you can see that we have declared a pointer of class type which points to class's object. We can access data members and member functions using pointer name with arrow -> symbol.
Pointers to members allow you to refer to nonstatic members of class objects. You cannot use a pointer to member to point to a static class member because the address of a static member is not associated with any particular object.
To access members of a structure through a pointer, use the arrow operator.
Like this:
(obj.*obj.pf)(0, 1);
Member access (.
) has a higher precedence than a pointer to member operator so this is equivalent to:
(obj.*(obj.pf))(0, 1);
Because function call also has higher precedence than a pointer to member operator, you can't do:
obj.*obj.pf(0, 1) /* or */ obj.*(obj.pf)(0, 1)
As that would be equivalent to:
obj.*(obj.pf(0, 1)) // grammar expects obj.pf to be a callable returning a
// pointer to member
pf is a method pointer, and you want to invoke the method it points to, so you have to use
(obj.*obj.pf)(1, 2);
It says the object obj you invoke the method pointed by pf
See result here :
http://ideone.com/p3a5G
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