typecasting in C

You can't typecast between and array and an int but you can copy the memory between them using memcpy. Something like...

memcpy(&i, send, sizeof(int));
memcpy(send, &i, sizeof(int));

http://www.cplusplus.com/reference/clibrary/cstring/memcpy/
void * memcpy ( void * destination, const void * source, size_t num );

Note, that the result you'll get from this will be dependent upon the endianess of the platform you are using. THis being the case this code is non-portable.
http://en.wikipedia.org/wiki/Endianness

Alternatively, use a union

union myUnion
{
   int i;
   char send[4];
};

http://www.cplusplus.com/doc/tutorial/other_data_types/

Note the C standard doesn't require an int to be 4 bytes so the code examples above are non-portable. This isn't likely to be a problem if you are only targeting a single platform but it is something you need to be aware of.

>> You can't typecast between and array
Actually, just to clarify... you can but the result won't be what you are expecting. Sorry I didn't mean that statement to be misleading... what you want to do isn't typecast but to convert and the semantics and the result are different.

actually i need to send a char msg ....I can send only char.... so I need to convert to the int to char...

actually i need to send that integer value...

ya  i need to convert to char....

and later I need to covert that char to integer again...

Ok, so I've shown you how to do that above {http:#24382735} ... do you need any more clarification? Is there something that doesn't make sense that maybe I can try and explain better for you?

no i could get ur point...

can i able to convert int to char  in C language....

>> can i able to convert int to char  in C language

You can assign an int to a char but this will lose data (a char normally being only 1 byte and an int 4 on most 32 bit platforms) so to prevent the compiler warning you it is necessary to use a cast.

int i = <somevalue>;
char c = (char) i;

int i = 20;
char send[4] = ?

Here what u want to store?
  Here 'Send' can hold 4 chars. But u are having only one int.
  So if u want to store one int only, u can use
               send[0] = (char) i;
Else
  evilrix has already explained well how to use UNION.

If u want all field of array should contain value of i, the can use
      int i = 20;
      char arr[4] = {(char)i, (char)i,(char)i, (char)i};
      OR //char arr[4] = {i, i, i, i};
//Botth are acceptable. but problem is if value of i is greater than 255 then it'd be trucated.There is also another logic

      for(int x = 0; x<4; ++x)
            printf("%c ", arr[x]);// If u want char to be printed. If want integer to be printed u can use %i instead of %c

Actually, if u can  explain ur need completely, like waht exactly u want, it'll help us to help u in a better way.  

shragi, do you need any more information on this? If so please ask away, otherwise please don't forget to close your question when you have suitable answers.