you are just changing the pointers . so second ptr will point to third
string and third will point to second string
On Wed, Sep 7, 2011 at 11:34 AM, Shachindra A C <sachindr...@gmail.com>wrote:
> I assume you expect to see the strings interchanged. But you are not
> changing anything in the memory. So in main(), pstr[0] and pstr[1] contains
> whatever was there earlier. If you change the parameters of the swap to
> swap(char **,char **), pass the addresses of pstr[0] ,pstr[1] and then do
> whatever u r doing in swap, the strings will get exchanged.
>
> On Wed, Sep 7, 2011 at 11:26 AM, piyush agarwal <pshagl...@gmail.com>wrote:
>
>> #include<stdio.h>void swap(char *, char *);
>> int main()
>> {
>> char *pstr[2] = {"Hello", "piyush"};
>> swap(pstr[0], pstr[1]);
>> printf("%s\n%s", pstr[0], pstr[1]);
>> return 0;
>> }void swap(char *t1, char *t2)
>> {
>> char *t;
>> t=t1;
>> t1=t2;
>> t2=t;
>> }
>>
>>
>>
>> --
>> Piyush Agarwal
>> Final Year Undergraduate
>> Department of Computer Engineering
>> Malaviya National Institute of Technology
>> Jaipur
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>
>
>
> --
> Regards,
> Shachindra A C
>
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