Asked  7 Months ago    Answers:  5   Viewed   69 times

How do I obtain a function pointer for a class member function, and later call that member function with a specific object? I’d like to write:

class Dog : Animal
{
    Dog ();
    void bark ();
}

…
Dog* pDog = new Dog ();
BarkFunction pBark = &Dog::bark;
(*pBark) (pDog);
…

Also, if possible, I’d like to invoke the constructor via a pointer as well:

NewAnimalFunction pNew = &Dog::Dog;
Animal* pAnimal = (*pNew)();    

Is this possible, and if so, what is the preferred way to do this?

 Answers

24

Read this for detail :

// 1 define a function pointer and initialize to NULL

int (TMyClass::*pt2ConstMember)(float, char, char) const = NULL;

// C++

class TMyClass
{
public:
   int DoIt(float a, char b, char c){ cout << "TMyClass::DoIt"<< endl; return a+b+c;};
   int DoMore(float a, char b, char c) const
         { cout << "TMyClass::DoMore" << endl; return a-b+c; };

   /* more of TMyClass */
};
pt2ConstMember = &TMyClass::DoIt; // note: <pt2Member> may also legally point to &DoMore

// Calling Function using Function Pointer

(*this.*pt2ConstMember)(12, 'a', 'b');
Tuesday, June 1, 2021
 
skrilled
answered 7 Months ago
11

Functions and function references (i.e. id-expressions of those types) decay into function pointers almost immediately, so the expressions func and f_ref actually become function pointers in your case. You can also call (***func)(5) and (******f_ref)(6) if you like.

It may be preferable to use function references in cases where you want the &-operator to work as though it had been applied to the function itself, e.g. &func is the same as &f_ref, but &f_ptr is something else.

Wednesday, June 9, 2021
 
braindamage
answered 6 Months ago
86

Danny Kalev explains this quite nicely:

The Underlying Representation of Pointers to Members

Although pointers to members behave like ordinary pointers, behind the scenes their representation is quite different. In fact, a pointer to member usually consists of a struct containing up to four fields in certain cases. This is because pointers to members have to support not only ordinary member functions, but also virtual member functions, member functions of objects that have multiple base classes, and member functions of virtual base classes. Thus, the simplest member function can be represented as a set of two pointers: one holding the physical memory address of the member function, and a second pointer that holds the this pointer. However, in cases like a virtual member function, multiple inheritance and virtual inheritance, the pointer to member must store additional information. Therefore, you can't cast pointers to members to ordinary pointers nor can you safely cast between pointers to members of different types.

To get a notion of how your compiler represents pointers to members, use the sizeof operator. In the following example, the sizes of a pointer to data member and a pointer to a member function are taken. As you can see, they have different sizes, hence, different representations:

struct A
{
 int x;
 void f();
};
int A::*pmi = &A::x;
void (A::*pmf)() = &A::f;
int n = sizeof (pmi); // 8 byte with my compiler
int m = sizeof (pmf); // 12 bytes with my compiler

Note that each of these pointers may have a different representation, depending on the class in question and whether the member function is virtual.

Saturday, July 3, 2021
 
Hilmi
answered 5 Months ago
43

Yes, it's guaranteed to work. From [expr.static.cast]:

A prvalue of type “pointer to member of D of type cv1 T” can be converted to a prvalue of type “pointer to member of B of type cv2 T”, where B is a base class (Clause 10) of D, if cv2 is the same cv-qualification as, or greater cv-qualification than, cv1.70 If no valid standard conversion from “pointer to member of B of type T” to “pointer to member of D of type T” exists (4.11), the program is ill-formed. The null member pointer value (4.11) is converted to the null member pointer value of the destination type. If class B contains the original member, or is a base or derived class of the class containing the original member, the resulting pointer to member points to the original member.

In this case, we're converting a pointer to member of CDerived of type void() to a poitner to member of CBase ov type void(). CBase is a base of the class containing the original member, so the resulting pointer points to the original member.

From [expr.mptr.oper]:

Abbreviating pm-expression.*cast-expression as E1.*E2, E1 is called the object expression. If the dynamic type of E1 does not contain the member to which E2 refers, the behavior is undefined.

In this case, pointer pointers to the original member. base has that member. So this is fine.


Note that in your example, base is actually a CDerived*. It would be just as valid to write:

CDerived d;
CBase* b = &d;

(b->*pointer)(); // ok - the dynamic type of b contains the member to which pointer refers
Sunday, October 10, 2021
 
Rasmus Puls
answered 2 Months ago
70

You have to instantiate c1 and f2. Try this:

public partial class Form1 : Form  
{
    Form2 frm2;
    Class1 cl;
    public Form1()  
    {  
        c1 = new Class1();
        InitializeComponent();  
    }
    private void button1_Click(object sender, EventArgs e)
    {
      frm2 = new Form2();
      cl.DoSomethingWithText(frm2);
      frm2.Show();
    } 
}

class Class1
{

    public void DoSomethingWithText(Test1.Form2 form)
    {
        form.richTextBox1.Text = "Blah blah blah";
    }
}

UPDATE

As Keith has pointed out, because you're instantiating a new version of Form2, the rich textbox will never show the blah blah blah code. I've updated the sample to fix this.

Sunday, November 14, 2021
 
Ripon Al Wasim
answered 2 Weeks ago
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