This is very bad. Change your compiler if it compiles this stuff :)

btw.. which compiler is it?

Output for me :
ro...@rohit-laptop:~/dump$ gcc c.c
c.c: In function ‘main’:
c.c:14: error: incompatible types when assigning to type ‘char[20]’ from
type ‘char *’
c.c:15: error: incompatible types when assigning to type ‘char[20]’ from
type ‘char *’

--------------------------------------------------
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14


On Sun, Jun 13, 2010 at 8:13 AM, Mahesh_JNU <[email protected]>wrote:

> Well
>
> As we know for copying the string we can can copy it as a simple variable
> as in case of address copying.
> when u r doing names[3] = names[4] , it means u r trying to copy it
> directly
> bt in the case of  char *names[] , as it is the array of pointers so u can
> copy the address from one pointer to another pointer....
>
> Thanks
>
>
> On Sat, Jun 12, 2010 at 9:12 PM, divya <[email protected]> wrote:
>
>> #include<stdio.h>
>>
>> int main()
>> { char names[][20]={
>> "roshni",
>> "manish",
>> "sona",
>> "baiju",
>> "ritu"
>> };
>> int i;
>> char *t;
>> t=names[3];
>> names[3]=names[4];
>> names[4]=t;
>> for(i=0;i<=4;i++)
>> printf("%s",names[i]);
>> printf("\n");
>> return 0;
>> }
>>
>> here i get l value required as error and if i replace char names[][2]
>> with char *names[].. then there is no error nd the names[3] n names[4]
>> interchange
>> plzzzz explain why???
>>
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>
>
> --
>   Mahesh Giri
>  MCA Final Sem
> JNU, New Delhi
>
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