http://www.daniweb.com/software-development/c/threads/339649
Thank you,
Sid.



On Wed, Sep 7, 2011 at 11:43 AM, siddharam suresh
<[email protected]> wrote:
> my mistake,
> for 2)you cant change the address of the any array,(use change it to
> char *t1="xyz",*t2="abc"; )
> i thought pstr is array of array of strings(but  is array of character
> pointer),
> @sachindraac is right
>
>  char *pstr[2] = {"Hello", "piyush"};
>
> Thank you,
> Sid.
>
>
> On Wed, Sep 7, 2011 at 11:34 AM, Shachindra A C <[email protected]> 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 <[email protected]> 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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