int *temp[5] would be read as "temp is an array of 5 pointers to int
and
int (*temp)[5] would be read as "temp is a pointer to an array of 5
integers.
not sure if you are looking for something else.
Thanks and Regards,
Naveen
Saquib Imam wrote:
Can u tel me the difference between
Int *temp[5];
And
Int (*temp)[5];
Thanks & Regards,
Saquib Imam
------------------------------------------------------------------------
*From:* kernelnewbies-bounce@xxxxxxxxxxxx
[mailto:kernelnewbies-bounce@xxxxxxxxxxxx] *On Behalf Of *sahlot arvind
*Sent:* Wednesday, December 26, 2007 10:34 AM
*To:* Saquib Imam
*Cc:* Cihan Kömeçoğlu; C_C_Kuo@xxxxxxxxxxxxxx; kernelnewbies@xxxxxxxxxxxx
*Subject:* Re: Question about double pointers assignment
>#include<stdio.h>
>#include<stdlib.h>
>int exterint[5][5];
>int exterint1[5];
>int main(void){
>int index0, index1;
>int **temp1;
>int *temp2;
>temp2=exterint1; //*****
>temp1 = exterint; //xxxxxxxxx
>The compiler will warn me that at line marked as xxxxx is "assignment
from incompatible pointer type."
> But "******" doesn't get any warning.
Here exterint1 is a pointer to the first element of one dimensional
array, which is nothing but an integer. Thus "exterint1" is nothing
but simply a pointer to an integer. Thats why you can assign it
"temp2" which is also a pointer of same type.
But keeping in mind the way two dimensional arrays are treated,
"exterint" is a pointer to an array of integers of size 5. Note that
it is not a pointer to pointer to an integer, while "temp1" is a
pointer to pointer to integer. Thus when in "temp1 = exterint" the
assignment is not correct and hence warning.
To remove it what you can do it, declare "temp1" as a pointer to an
array of integer of size 5. Something like this -
int (*temp1)[5];
temp1 = externint;
Hope it helps.
Best regards
Arvind Sahlot
On 12/26/07, *Saquib Imam* <saquib.imam@xxxxxxxxx
<mailto:saquib.imam@xxxxxxxxx>> wrote:
…
int exterint[5][5];
int exterint1[5];
…
int **temp1;
int *temp2;
….
temp2=exterint1; //*****
temp1 = exterint; //xxxxxxxxx "ERROR GIVING LINE"
In the above case exterint1 will be pointing to the first element of
the integer array, so it can be assigned to a integer pointer.
So the assignment
temp2=exterint1; is correct.
While in case of exterint , it will be pointing to the first element
to the first row , again its pointing to a integer not a pointer to
the integer while temp1 is a pointer to the pointer to an integer.
So the assignment
temp1 = exterint gives an error.
Thanks & Regards,
Saquib Imam
------------------------------------------------------------------------
*From:* kernelnewbies-bounce@xxxxxxxxxxxx
<mailto:kernelnewbies-bounce@xxxxxxxxxxxx> [mailto:
kernelnewbies-bounce@xxxxxxxxxxxx
<mailto:kernelnewbies-bounce@xxxxxxxxxxxx>] *On Behalf Of *Cihan
Kömeçoglu
*Sent:* Monday, December 24, 2007 4:26 PM
*To:* C_C_Kuo@xxxxxxxxxxxxxx <mailto:C_C_Kuo@xxxxxxxxxxxxxx>
*Cc:* kernelnewbies@xxxxxxxxxxxx <mailto:kernelnewbies@xxxxxxxxxxxx>
*Subject:* Re: Question about double pointers assignment
I think , the problem is there
temp1 = exterint
temp1 is pointer to pointer,not pointer to int but you assigned adress
of exterint1 and this array of integer; not array of pointer.
For example like this give you same warning:
int * temp1;
int a;
temp1 = a;
Warrning:"assignment from incompatible pointer type
This is correct if you do like below
int *temp1
int exterint[5];
temp1 = exterint;
Monday, December 24, 2007, 11:49:37 AM, you wrote:
*>*
Dear all:
I write a program like below:
#include<stdio.h>
#include<stdlib.h>
int exterint[5][5];
int exterint1[5];
int main(void){
int index0, index1;
int **temp1;
int *temp2;
temp2=exterint1; //*****
temp1 = exterint; //xxxxxxxxx
…..
…..}
The compiler will warn me that at line marked as xxxxx is "assignment
from incompatible pointer type."
But "******" doesn't get any warning.
Is there any restriction about assigning multi-layer array or
something about pointer I miss?
Appreciate your help,
cckuo
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