Pointers to members in C++

Question

#include <iostream>
using namespace std;

class Sample
{
    public:
        int *pxx;
        int x;

        void setD(int y)
        {
            x=y;
        }

        void print()
        {
            int Sample::*px = &Sample :: x; 
            cout<<"\nx : "<<x;
            cout<<"\nAddress of x : "<<&x;
            cout<<"\nValue of X indirected through px :"<<this->*px;
            //cout<<"\nValue of X indirected through px :"<<*px; ERROR
            //cout<<"\nAddress of x i.e. px : "<<px; NO ERROR BUT UNDESIRED OUTPUT(The output is most of the time '1')
            //cout<<"\nAddress of x i.e. px : "<<this->px; ERROR
        }
};

I've read that when a pointer is declared using the syntax data type <class_name> :: * <pointer> = &<class_name> :: <variable_name> it acts like a class member, then why am I not allowed to execute those statements given in the comments of the above program(except for 1). Is there any difference between px and pxx(except that of scope)?

you can refer to http://stackoverflow.com/questions/670734/c-pointer-to-class-data-member regarding how to use pointer to class member. — user258367, Aug 15 '15 at 06:08
I have referred to it before. I just want to know what is the scope of this _Pointer to a member_ and whether it classifies as a **class member** or not? — Zaid Khan, Aug 15 '15 at 07:17

AnT stands with Russia · Accepted Answer · 2015-08-15T08:46:40.937

3

Well, no it is not supposed to "act like a class member". That's just not true. I don't know where you got it.

It is supposed to act as a value of pointer-to-member type. And such pointers require ->* or .* operators to dereference

cout << "\nValue of X indirected through px :" << this->*px;
cout << "\nValue of X indirected through px :" << (*this).*px;

And, as you can see, these operators require a specific object on the left-hand side. You cannot just dereference px by itself. It has no meaning by itself.

A value of pointer-to-data-member type is a language-level implementation of the low-level concept of "offset" (offset of data measured from the beginning of the containing object). In order find the actual data through given offset, you have to have a starting point from which to measure that offset. That is what the object expression of the left-hand side of -> or .* represents.

edited Aug 15 '15 at 08:46

answered Aug 15 '15 at 06:03

AnT stands with Russia

312,472
42
525
765

I am not able to grasp "Pointer of Pointer-to-class-member"(Did you meant Pointer to Pointer/Double Pointer?). Can you elaborate on this please(with a simple example.) And why is the statement `cout<<"\nAddress of x i.e. px : "< – Zaid Khan Aug 15 '15 at 06:55
@Khan Zaid: "Pointer-to-member" is a special kind of entity, significantly different from a normal regular pointer. So, there are pointers of *regular kind* and there are pointers of *pointer-to-member kind*. On the second thought, maybe this redundant-sounding terminology should be avoided. One can say the same thing without it. (I'll try to rephrase the answer). And no, it has nothing to do with double pointers. – AnT stands with Russia Aug 15 '15 at 08:46
And why do you expect `cout << px;` to generate an error? This code does not attempt to *dereference* the pointer, meaning that what I said above about *dereferencing* such pointers does not apply. This code simply uses the fact that such pointers (just like regular pointers) can be converted to `bool` type. I.e. it is equivalent to `cout << (bool) px;` – AnT stands with Russia Aug 15 '15 at 08:49
Well don't these 'Pointers to Members' "have no meaning by itself", and then why does `cout<px;` generates an error then? Am I missing something? – Zaid Khan Aug 15 '15 at 09:07
@Khan Zaid: Well... There are valid expressions with pointers to members. And there are invalid expressions with pointers to members. `px`, `this->*px` and `(*this).*px` are valid expressions. `this->px` is an invalid one. Why does it surprise you? Why do you expect it compile? What should it do, in your opintion? – AnT stands with Russia Aug 15 '15 at 09:12
Well... If a pointer to a member really requires an object on its left side then `this->px` should be valid and `px` not? – Zaid Khan Aug 15 '15 at 09:17
@Khan Zaid: I never said that pointer-to-member requires an object on the LHS. I said: the operation of *dereferencing* a pointer-to-member requires an object on the LHS. Expression `px` does not *dereference* the pointer, so the LHS remark dos not apply. And expression `this->px` is simply invalid. – AnT stands with Russia Aug 15 '15 at 10:01
Sorry if my question sounds stupid, but why do we use this method, for example, why can't we use Method 1, code link [here.](http://justpaste.it/n30t) – Zaid Khan Aug 17 '15 at 14:11

Pointers to members in C++

1 Answers1