#include<stdio.h>
void swap(char **p,char **q)
{
char *t;
t=*p;
*p=*q;
*q=t;
}
int main(){
char *ps[2]={
"Hello",
"Good Morning",
};
swap( &ps[0] , &ps[1]);
printf("%s \t %s\n",ps[0],ps[1]);
return 0;
}
This one breaks the ice ... as you have to change the contents of the array
which is a pointer , you have to send a pointer to that pointer only that
pointer as if u do it will be taken as local variable and no effect will be
visible in actual array.
On Fri, Jun 17, 2011 at 3:17 AM, udit sharma <[email protected]>wrote:
> Ohh Sry... The qus was:
>
>
>> #include<stdio.h>void swap(char *,char *);int main(){char *ps[2]={
>>> "Hello",
>>> "Good Mornning",
>>> };swap(ps[0],ps[1]);printf("%s \t %s\n",ps[0],ps[1]);return 0;}
>>> void swap(char *p,char *q){char *t;t=p;p=q;q=t;}
>>>
>>>
>>> why the output is:
>>> Hello Good Mornning
>>>
>>>
>>>
>>>
>>>
>>>
>>> Regards
> UDIT
> DU- MCA
>
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Regards,
--
Rohit Sindhu.
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