@oppilas
char *ptr="hello"
in this case the string becomes constant but not the ptr, you can do this.
char *ptr="hello";
char arr[]="hi";
ptr[0]="B"; //not work
ptr=arr; //work
arr[0]="H"; //work
ptr[0]="N"; //work

Only string "Hello" becomes constant, it cant be changed.

Read - const char *ptr , char const *ptr, char * const ptr for more details.

I think you now got it.

On Sat, Jun 25, 2011 at 4:59 PM, oppilas . <jatka.oppimi...@gmail.com>wrote:

> Thanks all :).
>
>
> On 6/25/11, Anantha Krishnan <ananthakrishnan....@gmail.com> wrote:
> > When we declare *char *str="hello";*
> >
> > this "hello" will be stored in the read-only memory i.e *TEXT Segment*.
> >
> > so when we try to write the read-only memory by **str='w';* it will
> > throw *Segmentation
> > fault*.
> >
> > Obviously we must allocate some memory in heap to modify it like:
> > *char *str=(char *)malloc(1024);*
> >
> > Thanks & Regards,
> > Anantha Krishnan
> >
> > On Sat, Jun 25, 2011 at 2:16 PM, Adarsh <s.adars...@gmail.com> wrote:
> >
> >> char array[] = "hello";
> >> char *pointer = "hello";
> >>
> >> array is an array, enough to store sequence of characters and '\0'
> >> array will always refer to same storage.
> >> Here, pointer is initialized to point to a string constant, pointer
> >> may be modified, but you cannot chage string contents
> >>
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-- 
*Regards
Sagar Pareek
M.Tech cse (pur.)
Motilal Nehru National Institute Of Technology (MN NIT)
Allahabad*

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