Re: Conversion of pointer-to-member
On 7/26/07 9:20 PM, in article 5gsp63F3hnmd4U1@mid.dfncis.de, "Frank
Birbacher" <bloodymir.crap@gmx.net> wrote:
Hi!
I wonder why pointers to members seem so limited to me. I tried to do
the following:
struct Base {
virtual void doBar() =0;
};
template<typename T>
struct Dev : Base {
void doBar();
};
struct Test {
Dev<int> i;
Dev<float> f;
};
int main() {
Base Test::* const pt = &Test::i; //ERROR here
Test t;
(t.*pt).dobar();
}
It results in an error: a value of type "Dev<int> Test::*" cannot be
used to initialize an entity of type "Base Test::*const".
Well, sure there is an Instance of Base in every object of type Test.
Can I have a pointer-to-member to it?
No, because applying a member pointer to a class object obtains the member
object itself - not a polymorphic pointer or reference to the member object.
In practical terms, requiring the exact type of the member helps to prevent
the program from inadvertently "slicing" the member object (that is, copying
only the base class portion of an object of a derived class).
Furthermore, a program can always obtain a polymorphic pointer to the member
object once the member pointer has been applied to the class object:
Base * p = &(t.*pt);
A better idea I think would be to to add an routine to Test that returns a
polymorphic pointer of the desired type:
struct Test
{
Dev<int> i;
Dev<float> f;
Base* get() const { return &i; }
};
At least a member function pointer would provide better encapsulation than a
member data pointer (and in fact, member data pointers are seldom used in
C++ programming):
Base* (Test::*pmf)() = &Test::get;
Test t;
Base * bp = (t.*pmf)();
Greg
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