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Dave Rahardja
Guest
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 Posted: Sun Dec 19, 2004 8:12 pm Post subject: Returning ptr to a virtual member function of a base class |
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Consider the following example:
---
#include <iostream>
class Base
{
public:
virtual void fn() { std::cout << "Base::fn()n"; }
};
typedef void (Base::*Base_fn_ptr)();
class Derived: public Base
{
public:
virtual void fn() { std::cout << "Derived::fn()n"; }
static Base_fn_ptr Get_Base_Fn_Ptr() { return &Base::fn; }
};
int main()
{
Derived obj;
Base_fn_ptr ptr = Derived::Get_Base_Fn_Ptr();
(obj.*ptr)();
return 0;
}
---
Running the example causes "Derived::fn()" to be output. Is there a way
to craft Get_Base_Fn_Ptr() so that it returns the address of Base::fn()
instead of a virtual function-calling stub?
In other words, I want to defeat the polymorphic behavior of obj.
- Dave Rahardja
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GianGuz
Guest
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| Posted: Sun Dec 19, 2004 9:33 pm Post subject: Re: Returning ptr to a virtual member function of a base cla |
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If I understand what you need, the way to defeat the polimorfic
behaviour of obj is to use the explicit qualification mechanism in
calling fn.
In your code you only have to call:
obj.Base::fn(); // explicit call to Base fn()
obj.Derived::fn(); // explicit call to Derived fn()
Gianguglielmo
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Jonathan Mcdougall
Guest
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| Posted: Sun Dec 19, 2004 9:54 pm Post subject: Re: Returning ptr to a virtual member function of a base cla |
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Dave Rahardja wrote:
| Quote: | Consider the following example:
---
#include <iostream
class Base
{
public:
virtual void fn() { std::cout << "Base::fn()n"; }
};
typedef void (Base::*Base_fn_ptr)();
class Derived: public Base
{
public:
virtual void fn() { std::cout << "Derived::fn()n"; }
static Base_fn_ptr Get_Base_Fn_Ptr() { return &Base::fn; }
};
int main()
{
Derived obj;
Base_fn_ptr ptr = Derived::Get_Base_Fn_Ptr();
(obj.*ptr)();
return 0;
}
---
Running the example causes "Derived::fn()" to be output. Is there a way
to craft Get_Base_Fn_Ptr() so that it returns the address of Base::fn()
instead of a virtual function-calling stub?
In other words, I want to defeat the polymorphic behavior of obj.
|
By using pointer to member functions, no. This system was especially
made for these cases: to make sure wherever in the hierarchy you take an
address, the correct function gets called.
You are probably aware you can call it directly by qualifying the function :
obj->Base::fun();
However, look at that :
# include <iostream>
class Base
{
public:
void f()
{
std::cout << "in base::f()";
}
virtual void vf()
{
f();
}
};
class Derived : public Base
{
public:
virtual void vf()
{
std::cout << "in derived::vf()";
}
};
typedef void (Base::*Pf)();
Pf test1()
{
return &Base::f;
}
Pf test2()
{
return &Base::vf;
}
int main()
{
Base *b = new Derived;
Pf pf = test1();
(b->*pf)(); // in derived::vf()
pf = test2();
(b->*pf)(); // in base::vf()
}
This may or not be what you look for.
Jonathan
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Dave Rahardja
Guest
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| Posted: Mon Dec 20, 2004 4:50 am Post subject: Re: Returning ptr to a virtual member function of a base cla |
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GianGuz wrote:
| Quote: | If I understand what you need, the way to defeat the polimorfic
behaviour of obj is to use the explicit qualification mechanism in
calling fn.
In your code you only have to call:
obj.Base::fn(); // explicit call to Base fn()
obj.Derived::fn(); // explicit call to Derived fn()
Gianguglielmo
|
That's exactly what I'm looking for. Thanks! I guess the syntax of the
explicit call escaped me.
-dr
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