C / C++ / MFC

11-Mar-05 0:38

Hi,

I have a dialog app that has a list control that lists a dynamically set number of items. During the operation of the programme these items can become active or inactive but user is only allowed to set one item at a time. Hence I have a single selection list box.

During the operation of the programme there is always one item in use. And here in lies the problem...

...on startup how do you get a list control to select an item manually?

Thanks in advance,
Rich

"Programming today is a race between software engineers striving to build bigger and
better idiot-proff programs, and the Universe trying to produce bigger and better idiots.
So far the Universe is winning." -- Rich Cook

ThatsAlok11-Mar-05 2:01

ThatsAlok

11-Mar-05 2:01

Look for this Function
CListCtrl::SetItemState

[code]

for selecting
 [CListCtrl Object].SetItemState([Item Number], LVIS_SELECTED, LVIS_SELECTED);

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Alok Gupta
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RichardS11-Mar-05 3:26

11-Mar-05 3:26

I have been using this call that I have been using but calling this in InitDialog but with no luck. If I call it is OnClick (for example) then it works fine. Do you know when would be the best time to call it?

Rich

"Programming today is a race between software engineers striving to build bigger and
better idiot-proff programs, and the Universe trying to produce bigger and better idiots.
So far the Universe is winning." -- Rich Cook

Ravi Bhavnani11-Mar-05 4:30

Ravi Bhavnani

11-Mar-05 4:30

RichardS wrote:
Do you know when would be the best time to call it?

After the list control has been populated. You can certainly do this in OnInitDialog() or anywhere else for that matter.

/ravi

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C++ HELL : Object Addresses

ThatsAlok11-Mar-05 19:36

ThatsAlok

11-Mar-05 19:36

I hope, you have got your answer from Mr Ravi, now i believe nothing is left for writing here!!

[Vote One Here, Complete my Survey....]

Alok Gupta
visit me at http://www.thisisalok.tk

"I ~~Think~~ Believe this Will Help"

vikramlinux11-Mar-05 0:18

11-Mar-05 0:18

Hi Friends,
Please check following examples:
Example 1:
class Base1{
public:
int data0;
};

class Base2:public Base1{
public:
int data0;
};

class Derived: public Base2{
public:
int data0;
};

void main(void){
Derived D;
Base1 *bptr1= &D;
Base2 *bptr2=&D;
}

In above case :
Address of D=0x0012ff74;
Address of bptr1=0x0012ff74
Address of bptr2=0x0012ff74;

Now exmaple 2:

class Base1{
public:
int data0;
};

class Base2{
public:
int data0;
};

class Derived: public Base1,public Base2{
public:
int data0;
};

void main(void){
Derived D;
Base1 *bptr1= &D;
Base2 *bptr2=&D;
}

In above case :
Address of D=0x0012ff74;
Address of bptr1=0x0012ff74
Address of bptr2=0x0012ff78;

My question is why the base2 pointer has different address in above situation?

RichardS11-Mar-05 1:09

11-Mar-05 1:09

Hi vikrams,

In you first example the compiler will create the inheritance as:

base1
  |
base2
  |
Derived

Which means that the compiler only ever needs to know about one int data0 as it comes from base1.

In the second example the tree will be:

base1               base2
  |                   |
  ---------------------
             |
          Derived

Which means now int data0 could come from base1 OR base2 so you it needs to create 2 addresses.

cheers,
Rich

"Programming today is a race between software engineers striving to build bigger and
better idiot-proff programs, and the Universe trying to produce bigger and better idiots.
So far the Universe is winning." -- Rich Cook

vikramlinux11-Mar-05 1:15

11-Mar-05 1:15

See Data has to be there in all cases. It can not be overridden.U may conside a case where all there are having different data mebers.

RichardS11-Mar-05 3:34

11-Mar-05 3:34

You are quite right but you missed one thing. You have not created different data members. What you have done is to create each instance of the class with the same data member. Thus the compiler just allocs the same memory place for it. Think of it like what a union does.

The data is there in all cases (base1::data0, base2::data0 and derived::data0 will be all valid calls) the compiler has just seen that the same data member is inherited and thus the same memory can be used. In the first eg this is no ambiguity between the where to locate the memory. In the second one there is, so two locations are located (the first inherited object will line up with the derived class).

Rich

"Programming today is a race between software engineers striving to build bigger and
better idiot-proff programs, and the Universe trying to produce bigger and better idiots.
So far the Universe is winning." -- Rich Cook

jan larsen11-Mar-05 1:14

jan larsen

11-Mar-05 1:14

Very simple ( Smile | :)

):
In your first example you're using simple inheritance, and in the second one you're using multiple inheritance.
Using simple inheritance, you'll have a simple object with no ambigous methods, methods of the same name are either hidden or overridden.
In the multiple inheritance situation in example two, you have a method name conflict.

The details about how this applies to the vtables and whatnot is a question for more able people than me Big Grin | :-D

"After all it's just text at the end of the day. - Colin Davies

"For example, when a VB programmer comes to my house, they may say 'does your pool need cleaning, sir ?' " - Christian Graus

vikramlinux11-Mar-05 1:20

11-Mar-05 1:20

See imagene a case where no methods for both type of inheritance. And anyways methods are always outside class objects.

jan larsen13-Mar-05 22:13

jan larsen

13-Mar-05 22:13

Que?

vikrams wrote:
And anyways methods are always outside class objects
Yes and no, because the instance will contain the vtable.

"After all it's just text at the end of the day. - Colin Davies

"For example, when a VB programmer comes to my house, they may say 'does your pool need cleaning, sir ?' " - Christian Graus

vikramlinux13-Mar-05 22:16

13-Mar-05 22:16

See don't consider a case of virtual function and inheritance in this case. I just want a clean idea what is happening behind the scene.

TheGreatAndPowerfulOz11-Mar-05 16:27

TheGreatAndPowerfulOz

11-Mar-05 16:27

What you've basically got here is something that is essentially the same as the following: (note this is not inheritance, but aggregation --- done to illustrate a point)

struct Derived
{
    Base1 b1;
    Base2 b2;
}

void main()
{
  Derived D;
  Base1* bptr1 = &D.b1;
  Base2* bptr2 = &D.b2;
}

The reason (Base1*)&D != (Base2*)&D is that each base class has to occupy some part of derived class Derived. In other words, Derived is composed of Base1 and Base2, so when you get a pointer to each of the base classes, it will point at the part of D that is that base class. Try switching the order of the base classes, you will see that the pointer values switch. That will illustrate that pointers to each of the base classes point to the part of Derived that is made up by the respective base class.

Re: windows.h / MessageBox

jan larsen13-Mar-05 22:16

jan larsen

13-Mar-05 22:16

A very good explanation.

"After all it's just text at the end of the day. - Colin Davies

"For example, when a VB programmer comes to my house, they may say 'does your pool need cleaning, sir ?' " - Christian Graus

windows.h / MessageBox

doneirik11-Mar-05 0:15

doneirik

11-Mar-05 0:15

hi ...posted a question yesterday concerning a problem with "MessageBox"

Now I know where the problem is, but I don´t know how to solve it.

Making a Windows Form App with VC++ .Net 2003, I add my own header file to the project. Trying to use MessageBox::Show("String") gives me an errormessage but only as long as the header file windows.h is included in the project.

Is this windows.h header file incompatible with VC++ or is there any tricks I have to know?

doneirik

vikramlinux11-Mar-05 0:53