Re: Function pointer to a method and use in STL

From:

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Mon, 6 Jul 2009 16:35:31 CST

Message-ID:

<e540a485-e97b-468b-9bdd-5e58889de1d5@q11g2000yqi.googlegroups.com>

On 6 Jul., 19:31, Verictor <stehu...@gmail.com> wrote:

Hi,

Given this implementation:

struct A
{
      static void something(int) {} // declare a static
method
}

void foo()
{
      void (*psomething)(int) = A :: something; // assign it
      void (*psomething)(int) = &A::something; // another way of
assigning it

      // calling the method
     (*psomething)(1);

}

The two assigning methods are the same except one is using
A::something and the other uses &A::something. In STL functors,
&A::something is always used. I wonder if there is particular reason?
Or does the ampersand in front of A really offer anything?

The answer has two aspects of functions two consider:

Static member functions are equivalent to free (non-member)
functions, but are hosted in the class scope. So the type
of A::something(int) is void(int) and we can take a normal
function pointer as we do for other free functions:

  A::something;
  &A::something;

is equivalent to given

void free_func(int);

and getting a function pointer by writing either of

  free_func;
  &free_func;

The result of all four expressions is

"pointer to void(int)"

If you mention now "STL functors", I assume you mean
something like std::mem_fun, right? (Note that with
std::ptr_fun your assertion doesn't hold: std::ptr_fun
exists to bind normal free function pointers, so
either of static member functions or free functions))

The declaration of std::mem_fun is

template<class S, class T>
mem_fun_t<S,T> mem_fun(S (T::*f)());

and this function accepts *not* a function pointer,
but a so-called "pointer to member function".
Such thingee is completely different from a
function pointer and a simplified description
of such a "pointer-to-member" is that it describes
some kind of offset of the member in the
corresponding class, which is *no* address in
the usual sense.

To get a pointer to member you need a non-static
member of a class and you need to apply the
construction:

&ClassName::MemberName

like in the following example:

struct S {
  void foo(int);
};

void bar() {
  &S::bar;
}

The result of the expression &S::bar is:

"pointer to member of class S of type void(int)"

and note well the difference to the type of
the free functions mentioned above.

In case of "pointer-to-member" neither the
class-qualifier (not even inside the class) nor
the & operator are optional. [expr.unary.op]/4
is quite clear regarding this:

"A pointer to member is only formed when an
explicit & is used and its operand is a qualified-id
not enclosed in parentheses. [ Note: that is, the
expression &(qualified-id), where the qualified-id
is enclosed in parentheses, does not form an
expression of type ?pointer to member.? Neither does
qualified-id, because there is no implicit conversion
from a qualified-id for a non-static member function
to the type ?pointer to member function? as there is
from an lvalue of function type to the type ?pointer
to function? (4.3). Nor is &unqualified-id a pointer
to member, even within the scope of the unqualified-
id?s class. ?end note ]"

HTH & Greetings from Bremen,

Daniel Kr?gler

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