The function swap just swaps it's local copy of pointers which does not mean
that it swap array elements. You have to do it explicitly.
On Fri, Jun 17, 2011 at 3:17 AM, udit sharma <sharmaudit...@gmail.com>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
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Rohit Sindhu
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