@rajeev: your code gives compilation error.
On Fri, Jul 29, 2011 at 11:39 PM, Ankur Khurana <ankur.kkhur...@gmail.com>wrote:
> no ,
> n[0] is *(n+0)
> so actually n is being dereferenced here .Check the basci diff
>
> p is a pointer , but n is a pointer to a pointer.
>
>
> On Fri, Jul 29, 2011 at 11:34 PM, Arshad Alam <alam3...@gmail.com> wrote:
>
>> wow great... but why it is so yaar?
>> ptr=n and ptr=n[0] is same na?
>>
>>
>> On Fri, Jul 29, 2011 at 11:27 PM, rajeev bharshetty <rajeevr...@gmail.com
>> > wrote:
>>
>>> #include<stdio.h>
>>>
>>> void main()
>>> {
>>> int n[3][3]= {
>>> 2,4,3,
>>> 6,8,5,
>>> 3,5,1
>>> };
>>> int i,*ptr;
>>> ptr= n;
>>> for(i=0;i<=8;i++)
>>> printf("\n%d",*(ptr+i));
>>>
>>> }
>>>
>>> In gcc 4.3.2 no error ,it is just showing a warning as
>>>
>>> ms50.c: In function ‘main’:
>>> ms50.c:11:8: warning: assignment from incompatible pointer type
>>>
>>> change the statement as ptr = n[0] warning vanishes
>>>
>>> On Fri, Jul 29, 2011 at 11:17 PM, Arshad Alam <alam3...@gmail.com>wrote:
>>>
>>>> what's the problem with line number 12?
>>>>
>>>> 1 #include<stdio.h>
>>>> 2 #include<conio.h>
>>>> 3 void main()
>>>> 4 {
>>>> 5 clrscr();
>>>> 6 int n[3][3]= {
>>>> 7 2,4,3,
>>>> 8 6,8,5,
>>>> 9 3,5,1
>>>> 10 };
>>>> 11 int i,*ptr;
>>>> 12 ptr=n;
>>>> 13 for(i=0;i<=8;i++)
>>>> 14 printf("\n%d",*(ptr+i));
>>>> 15 getch();
>>>> }
>>>>
>>>> --
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>>>
>>>
>>>
>>> --
>>> Regards
>>> Rajeev N B <http://www.opensourcemania.co.cc>
>>>
>>> "*Winners Don't do Different things , they do things Differently"*
>>>
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>>
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>
>
>
> --
> Ankur Khurana
> Computer Science
> Netaji Subhas Institute Of Technology
> Delhi.
>
>
--
Ankur Khurana
Computer Science
Netaji Subhas Institute Of Technology
Delhi.
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