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llewelly
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 Posted: Tue Aug 26, 2003 6:54 pm Post subject: Re: using member function pointers |
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[email]set.me (AT) free (DOT) fr[/email] (TB) writes:
| Quote: | I'm a bit confused here...
Given
class Base { ... public: virtual void foo(); ... }
class Derived : public Base { ... public: virtual void foo(); ... }
and
template <class A> void callMethod(A *a, void (A::*f)())
{ (a->*f)(); }
if you have
Base *p = new Derived;
callMethod(p, Base::foo);
what gets called - Base::foo or Derived::foo?
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Derived::foo.
| Quote: | In other words,
does this follow static or dynamic linkage.
[snip] |
Both. callMethod is instantiated on the static type of p, which is B*,
resulting in callMethod<B>(B *a, void (B::*f)()) . Within
callMethod<B>, (a->*f)() results in a virtual function call,
because Base::foo is a virtual function, and its address was
copied into f when callMethod<B> was called.
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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Ben Hutchings
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| Posted: Tue Aug 26, 2003 7:07 pm Post subject: Re: using member function pointers |
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In article <247c59f2.0308251133.1dff0cff (AT) posting (DOT) google.com>, TB wrote:
| Quote: | I'm a bit confused here...
Given
class Base { ... public: virtual void foo(); ... }
class Derived : public Base { ... public: virtual void foo(); ... }
and
template <class A> void callMethod(A *a, void (A::*f)())
{ (a->*f)(); }
if you have
Base *p = new Derived;
callMethod(p, Base::foo);
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An "&" is required before "Base::foo".
| Quote: | what gets called - Base::foo or Derived::foo? In other words,
does this follow static or dynamic linkage.
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What you're asking about is properly called binding, not linkage. The
answer is that a non-static member function pointer to a virtual member
function is dynamically bound, so Derived::foo is called.
| Quote: | If it follows static linkage, how can you get it to follow dynamic
linkage? And/or vice-versa?
snip |
It's not possible, so far as I know.
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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TB
Guest
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| Posted: Wed Aug 27, 2003 9:29 am Post subject: Re: using member function pointers |
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llewelly <llewelly.at (AT) xmission (DOT) dot.com> wrote
| Quote: | set.me (AT) free (DOT) fr (TB) writes:
I'm a bit confused here...
Given
class Base { ... public: virtual void foo(); ... }
class Derived : public Base { ... public: virtual void foo(); ... }
and
template <class A> void callMethod(A *a, void (A::*f)())
{ (a->*f)(); }
if you have
Base *p = new Derived;
callMethod(p, Base::foo);
what gets called - Base::foo or Derived::foo?
Derived::foo.
In other words,
does this follow static or dynamic linkage.
[snip]
Both. callMethod is instantiated on the static type of p, which is B*,
resulting in callMethod<B>(B *a, void (B::*f)()) . Within
callMethod<B>, (a->*f)() results in a virtual function call,
because Base::foo is a virtual function, and its address was
copied into f when callMethod<B> was called.
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So if I want callMethod to invoke Base's version (if I'm being really
twisted), how would I do it? By writing (a->Base::*f)() instead? Or
is there no way of invoking Base::foo on p in this way?
Thanks again,
TB
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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