Simple pointer object error C++

Question

I made a small program in c++ to learn more about pointer objects. However I keep getting an error that the pointer object "bar" is not a part of the class more specifically:

[Error] 'class ux' has no member named 'bar'

As shown below:

class ux {
private:
    int width;
    int height;

public:
    ux(int, int);
    ~ux();
    int area();
};

ux::ux(int a, int b)
{
    width = a;
    height = b;
}
ux::~ux(){}

int ux::area()
{
    return (width * height);
}

int main(int argc, char** argv) {
    ux * foo;
    ux * bar;

    foo = new ux(2, 2);
    foo->bar = NULL;
    cout << foo->area() << "\n";
    cout << foo->bar << "\n";   
}

Why is this happening? I know it's possible to point to another pointer object with the arrow operator but it is not working here.

---------------------Update-----------------------

I am not trying to do anything with foo->bar I just want to see the address or value printed out but I know that a pointer obj can point to another pointer obj. If someone can please specify a link that shows this technique in use I would greatly appreciate that.

What are you trying to do with foo->bar? In this case, foo is an instance of the ux class. ux doesn't have anything in it named "bar", which is why you're getting the error. — Moon
– Moon, Commented Nov 23, 2015 at 23:04
well, ux doesn't have a member called bar - so what did you expect foo->bar should do? — MikeMB
– MikeMB, Commented Nov 23, 2015 at 23:05
To learn about pointers, using new is not a requirement. foo ux(2, 2); foo* fptr = &ux; Then practice how to use pointers using fptr. — PaulMcKenzie
– PaulMcKenzie, Commented Nov 23, 2015 at 23:11
I think you do not understand what -> means. What you wrote is just "access a bar field in the object pointer by foo"... — Johan
– Johan, Commented Nov 23, 2015 at 23:11
it is hard for SO to answer your question since its not clear what you are trying to do. DO something simple first - then try to get clever. Use objects without pointers for example — pm100
– pm100, Commented Nov 23, 2015 at 23:17

Mark Ransom · Accepted Answer · 2015-11-23 23:53:34Z

1

The -> isn't used to access arbitrary pointers, it's used to access members of the class. The members of your ux class are width, height, and area. bar is a completely separate variable of type ux, and you would use it e.g. bar->width, bar->height, or bar->area().

answered Nov 23, 2015 at 23:53

Mark Ransom

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Nikola Butigan · Accepted Answer · 2015-11-23 23:26:47Z

0

In line foo->bar you are giving an order to object foo to do function from class ux called bar . Your defined class ux does not have attribute/function defined in class. You should read this article about operators .Its maybe easier to learn more about C++ Classes and objects then dive into pointers.

answered Nov 23, 2015 at 23:26

Nikola Butigan

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galdin · Accepted Answer · 2015-11-23 23:53:16Z

0

I know it's possible to point to another pointer object with the arrow operator but it is not working here.

Yes it is possible. The right way to do it is:

foo = bar;

Since foo is a pointer, it is essentially a variable with an address as its value. To make another variable point to the same object, all you have to do is assign the address to the other variable.

As a side note, remember that every new should have a delete.

answered Nov 23, 2015 at 23:53

galdin

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