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cbull
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 Posted: Fri Feb 24, 2006 3:06 am Post subject: Casting pointer to member. |
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class A
{
};
class B: public A
{
void f();
};
typedef void (A::*MPA)(void);
MPA mpA;
B b;
mpA = (MPA) &B::f;
b->*mpA();
Will it work? If not then when it will fail? I've heard something about
multi inheritance.
Tomasz
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++Hector C
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| Posted: Fri Feb 24, 2006 10:06 am Post subject: Re: Casting pointer to member. |
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I can see two reasons why it doesn't work.
1. Precedence: Function call () has higher precedence than pointer to
member object and pointer .* and ->*. You should use parenthesis.
2. The pointer to member pointer requires a pointer. You should use
pointer to member object instead.
The following code compiles (although you destroyed type safety with
the casting)
class A {};
class B : public A
{
public:
void f(){}
};
typedef void ( A::* MPA )( void );
int main()
{
MPA mpA;
B b;
mpA = ( MPA ) & B::f;
(b.*mpA)();
return 0;
}
++Hector C
cbull wrote:
| Quote: | class A
{
};
class B: public A
{
void f();
};
typedef void (A::*MPA)(void);
MPA mpA;
B b;
mpA = (MPA) &B::f;
b->*mpA();
Will it work? If not then when it will fail? I've heard something about
multi inheritance.
Tomasz
|
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kanze
Guest
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| Posted: Fri Feb 24, 2006 11:06 am Post subject: Re: Casting pointer to member. |
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cbull wrote:
| Quote: | class A
{
};
class B: public A
{
void f();
};
typedef void (A::*MPA)(void);
MPA mpA;
B b;
mpA = (MPA) &B::f;
b->*mpA();
Will it work?
|
It shouldn't even compile. The left operand of ->* must be a
pointer (barring user defined overloading), and b is not a
pointer.
There's also a problem of precedence.
| Quote: | If not then when it will fail? I've heard something about
multi inheritance.
|
If you replace b->*mpA() by (b.*mpA)(), the standard guarantees
that it will work. If you have a pointer p of type A*, and the
actual object type is B, (p->*mpA)() is also guaranteed to work.
I wouldn't be too surprised if some compilers got it wrong,
however. It isn't exactly the most used feature of C++.
--
James Kanze GABI Software
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34
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Bob Hairgrove
Guest
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| Posted: Fri Feb 24, 2006 11:06 am Post subject: Re: Casting pointer to member. |
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On 23 Feb 2006 21:28:50 -0500, "cbull" <cbull (AT) poczta (DOT) onet.pl> wrote:
| Quote: | class A
{
};
class B: public A
{
void f();
};
typedef void (A::*MPA)(void);
MPA mpA;
B b;
mpA = (MPA) &B::f;
b->*mpA();
|
You need parentheses here:
(b->*mpA)();
| Quote: | Will it work? If not then when it will fail? I've heard something about
multi inheritance.
Tomasz
|
As long as the dynamic type of the object used to dereference the
pointer is B, it should work. Otherwise, the behavior is undefined
(see section 5.2.9, paragraph 9 of the C++ standard). You should use
static_cast instead of a C-style cast, though.
--
Bob Hairgrove
NoSpamPlease (AT) Home (DOT) com
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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Bob Hairgrove
Guest
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| Posted: Fri Feb 24, 2006 3:06 pm Post subject: Re: Casting pointer to member. |
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On 24 Feb 2006 05:39:15 -0500, Bob Hairgrove <invalid (AT) bigfoot (DOT) com>
wrote:
Oops ... surely you meant:
B *b;
Of course, then you would need to initialize the pointer to some valid
object ... so maybe you didn't mean that?
| Quote: | mpA = (MPA) &B::f;
b->*mpA();
You need parentheses here:
(b->*mpA)();
|
Only if b is a B* and not a B will that work.
--
Bob Hairgrove
NoSpamPlease (AT) Home (DOT) com
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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Kodt
Guest
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| Posted: Fri Feb 24, 2006 3:06 pm Post subject: Re: Casting pointer to member. |
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|
cbull wrote:
| Quote: | class A
{
};
class B: public A
{
void f();
};
typedef void (A::*MPA)(void);
MPA mpA;
B b;
mpA = (MPA) &B::f;
b->*mpA();
Will it work? If not then when it will fail? I've heard something about
multi inheritance.
|
Syntax errors:
- using ->* instead of .*
- precedence of ()
should be fixed as follows
(b.*mpA)();
A pointer-to-member-of-T should be appliable to _any_ object of class T
(or its descendants).
It doesn't matter what compilers do in background (taking offsets and
virtualness into account): just imagine you substitute your value
manually.
A member function of B cannot be applied to an instance of A. Thus,
(MPA) &B::f is at least meaningless.
But downcasting (from member of A to pointer-to-member-of-B) is
reasonable, as you can invoke inherited functions.
In fact, your compiler does here a reinterpret_cast<MPA*> and, luckily,
MPA and MPB have same layout (just pointers to code) and same
background processing (nothing) - that's why your program work.
We can invent a case when it fail to work. For instance,
struct A {}; // no virtual functions
typedef void (A::*MPA)(void);
struct B : A { virtual ~B(); }; // let's introduce virtual functions
typedef void (B::*MPB)(void); // sizeof(MPB) > sizeof(MPA);
struct A2 {};
struct C : A, A2 {}; // multiple inheritance
typedef void (C::*MPC)(void); // sizeof(MPC) > sizeof(MPA);
So, conversion MPB->MPA is impossible without loss of data.
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