Function pointer

[George2]

Hello everyone,
In the following statements,

template <class R, class Tclass mem_fun_t : public unary_function
<T*, R>
 
{
 
R (T::*pmf)()
....
}
 

1. I think pmf is a type of function pointer, the return type of the
function is R and the function is a member function of type (class) T.
Is my understanding correct?

2. If yes, what is the parameter list of the function? Empty parameter
list?

3. I doubt whether it is useful to define a function pointer with
empty parameter list -- too restricted.
thanks in advance,
George

[Rolf Magnus]

George2 wrote:

Hello everyone,
In the following statements,

template <class R, class Tclass mem_fun_t : public unary_function
<T*, R>

{

R (T::*pmf)()
...
}
 

1. I think pmf is a type of function pointer,

With a semicolon added to the end of that line, pmf would be the name of a
member variable of mem_fun_t.

the return type of the function is R and the function is a member function
of type (class) T.

Yes.

2. If yes, what is the parameter list of the function? Empty parameter
list?

Yes.

3. I doubt whether it is useful to define a function pointer with
empty parameter list -- too restricted.

Well, it is restricted to functions that have exactly 0 parameters. If you
defined a pointer to function with 10 parameters, it would be restricted to
functions that take exactly 10 parameters.

[Michael DOUBEZ]

George2 a écrit :

Hello everyone,
In the following statements,

template <class R, class Tclass mem_fun_t : public unary_function
<T*, R>

{

R (T::*pmf)()
...
}
 

1. I think pmf is a type of function pointer, the return type of the
function is R and the function is a member function of type (class) T.
Is my understanding correct?

yes.
In brief it iss a pointer on a member fucntion of class T.

>
2. If yes, what is the parameter list of the function? Empty parameter
list?

Yes. No parameters.

>
3. I doubt whether it is useful to define a function pointer with
empty parameter list -- too restricted.

It is useful if you want to call a method of a class that takes no
parameters.

struct foo
{
static int gcounter=0;

foo(){id_=gcoun ter++;}

void bar(){std::cout <<"My id is "<<id_<<std::en dl;}

int id_;
};
std::vector<foo v(10);

std::for_each(v .begin(),v.end( ),std::mem_fun_ ref(&foo::bar)) ;
Of course, you could put parameters in a class and make it a wrapper
class of a more elaborate functor.

Another example from SGI:
struct B {
virtual void print() = 0;
};

struct D1 : public B {
void print() { cout << "I'm a D1" << endl; }
};

struct D2 : public B {
void print() { cout << "I'm a D2" << endl; }
};

int main()
{
vector<B*V;

V.push_back(new D1);
V.push_back(new D2);
V.push_back(new D2);
V.push_back(new D1);

for_each(V.begi n(), V.end(), mem_fun(&B::pri nt));
}

Michael

[Pete Becker]

On 2007-12-10 06:48:11 -0500, George2 <ge************ *@yahoo.comsaid :

Hello everyone,
In the following statements,

template <class R, class Tclass mem_fun_t : public unary_function
<T*, R>

{

R (T::*pmf)()
...
}
 

1. I think pmf is a type of function pointer,

No. It's a pointer to member function, which is not at all like a
pointer to (non-member) function.

the return type of the
function is R and the function is a member function of type (class) T.
Is my understanding correct?

Right.

>
2. If yes, what is the parameter list of the function? Empty parameter
list?

Yes.

>
3. I doubt whether it is useful to define a function pointer with
empty parameter list -- too restricted.

Read about what mem_fun_t is used for. The definition of pmf is exactly
what's needed.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)