C / C++ / MFC

Laxman Auti6-Jun-06 20:44

6-Jun-06 20:44

paras_s12 wrote:
ya u give solution but that is not solve my problem
i want to disable it from menu which is on first dailog box and i want to disable edit control on second dailog box so enablewindow demand of object and i declare object of that edit in second dialog box and in first it will not take that one

Create the pointer of the Secoend dialog box into first one and access the elements/Objects through that pointer.

Hope i understood your problem.

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p_6-Jun-06 21:21

6-Jun-06 21:21

\\\\\\\\\\\\\\\\\\\plz help me \\\\\\\\\\\\\\\\\\

i made a class of the the first dailog box which contain menu as CAfterone
and made a class of second dialog box which contain edit boxes as CAgaindialog now i declare
CAgaindialog *pop; in the header file of the CAfterone
ass i declare it in CAfterone :: public CDialog (public mode)
////////////can i declare it any where or in the class
CAfterone :: public CDialog (public mode)////////////

then i use to call second dailog box from menu as
void CPasswordDlg::OnDel()
{
d.DoModal();
pop->GetDlgItem(IDC_EDIT)->EnableWindow(FALSE);
}
but it will not do job for me

hi

-- modified at 3:43 Wednesday 7th June, 2006

Laxman Auti6-Jun-06 21:57

6-Jun-06 21:57

paras_s12 wrote:
but it will not do job for me

Try to make the Variable as extern , initialise it properly before invoking the enablewindow api.

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p_6-Jun-06 22:20

6-Jun-06 22:20

as i make variable of edit control of the type CEdit how could i make it extern
d.DoModal();
pop->GetDlgItem(m_nemp)->EnableWindow(FALSE);
i take m_nemp as CEdit type and call it but it will give error that undefind in class one

hi

Laxman Auti6-Jun-06 22:35

pointers to an array and array of pointers

6-Jun-06 22:35

wrote:
as i make variable of edit control of the type CEdit how could i make it extern
d.DoModal();
pop->GetDlgItem(m_nemp)->EnableWindow(FALSE);
i take m_nemp as CEdit type and call it but it will give error that undefind in class one

I mean make extern variable of the second dialog box.
access it after initialisation.

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Laxman Auti6-Jun-06 20:05

Re: pointers to an array and array of pointers [modified]

6-Jun-06 20:05

Hi gurus,
what is the difference between following statements

int *p[10];
int (*p)[10];

what is the memory structure for both?
How to access the elements?

Thanks in advance..

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_AnsHUMAN_ 6-Jun-06 20:21

_AnsHUMAN_

6-Jun-06 20:21

int *p[10];
it is a two dimensional array and its elements can be accessed as
p[0][0],p[0][1].....

Somethings seem HARD to do, until we know how to do them.
Wink | ;-)

_AnShUmAn_

-- modified at 2:25 Wednesday 7th June, 2006

Re: pointers to an array and array of pointers [modified]

Laxman Auti6-Jun-06 20:27

Re: pointers to an array and array of pointers [modified]

6-Jun-06 20:27

AnShUmAn wrote:
int *p[10];
it is a two dimensional array and its elements can be accessed as
p[0][0],p[0][1].....

i think int *p[10] is array of 10 pointers.. am i right?
where we have to allocate the memory for each..

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-- modified at 2:27 Wednesday 7th June, 2006

Hamid_RT6-Jun-06 21:03

Hamid_RT

6-Jun-06 21:03

See for int *p[10];

<br />
int *p[2];<br />
memset(&p,0,sizeof(int[2]));<br />
<br />
int b=10;<br />
p[0]=&b;<br />
int c;<br />
c=20;<br />
p[1]=&c;<br />
<br />
int A;<br />
A=*p[0];<br />
A=*p[1];<br />

whitesky

yang__lee6-Jun-06 20:48

yang__lee

6-Jun-06 20:48

int *p[10]
is a array of 10 pointers.

and i think second one is pointer to an array of 10 integers.

or you may declare it as int **p
assign memory to *p of 10 blocks so it will be an array
and then **p will be the pointer to array of 10 integers.

Leya

Viorel.6-Jun-06 21:29

Viorel.

6-Jun-06 21:29

Since in operator precedence the "[]" operator has a priority above the "*" operator, the expression "int *p[10]" denotes an array of ten pointers, each pointing to an integer. In "int (*p)[10]" the order of interpretation was changed, and here we have a pointer to an array of ten integers.

In the first case, the size of p is 40 bytes (in our compiler), in the second case, the size is 4 bytes.

In the first case, you can assign new value for each array element like this:

p[2] = &z;
p[7] = new int;

Expression like *p[2] here can be used to change the value of z:

*p[2] = 100;

In the second case, you first have to allocate an array and assign its address to the pointer:

p = (int (*)[10]) new int [10];

Now you can access elements from this array:

(*p)[3] = 100;
(*p)[5] = 200;

Then you can delete the array:

delete [] p;

Laxman Auti6-Jun-06 23:35

6-Jun-06 23:35

Viorel wrote:
p = (int (*)[10]) new int [10];

Is it necessary to typecast the allocated memory ?? Confused | :confused:

As per my knowledge the new operator returns the appropriate pointer.

or Should we allocate memory as
p = new int [5][10];

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Viorel.7-Jun-06 0:19

Viorel.

7-Jun-06 0:19

I think, with older compilers an assignment like "int (*p)[10] = new int [10]" would work, but today, in VS 2003 or GCC, it is not acceptable. Compiler complains: "cannot convert from 'int *' to 'int (*)[10]'". I suppose this is made to follow standards.

Regarding the second question, yes, you can make the assignment "int (*p)[10] = new int [5][10]" if you are going to interpret p as a pointer to bi-dimensional array. You can refer the elements like p[2][7].

Laxman Auti7-Jun-06 0:31

7-Jun-06 0:31

Viorel wrote:
int (*p)[10] = new int [10]

fine..

int (*p)[10] =(int (*)[10]) new int [10];

How many elements array able to hold.

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Viorel.7-Jun-06 1:18

Viorel.

7-Jun-06 1:18

As I know the number of array’s elements, as well as the number of dimensions in multi-dimensional array is not limited by the language, but is limited by the compiler, libraries and available resources.

The size of available memory is a factor which determines the array’s limits. If no enough memory for new operator, an exception is thrown.

Compiler and libraries included in VS 2003 enforce the total size of an array to be less then 0xFFFFFFFF. Therefore an array of integers cannot have more then approx. 1e9 elements, but the available memory space is usually smaller.

Laxman Auti6-Jun-06 20:00

6-Jun-06 20:00

RahulRKulkarni wrote:
I want to build an application of login and call google user.

I am going to use libjingle application.

any suggetions, ideas needed.

Search for google API's and through them validate the Google user.
following link have sample requests for validating the client login..

Here[^]

Hope this helps you.

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-- modified at 2:08 Wednesday 7th June, 2006

Rahul.RK6-Jun-06 20:10

Rahul.RK

6-Jun-06 20:10

Link give page not found error

Rahul Kulkarni

Laxman Auti6-Jun-06 20:20

6-Jun-06 20:20

RahulRKulkarni wrote:
Link give page not found error

Check it again... sorry some text missed in link

welcome to CP

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