@programming love:I can understand what u say but my doubt is that for the
first output which is 2, according to your example, p has address 10 which
it points to.And as u say int * tends to dereference 2 bytes so that wud be
now 10 and 11.finally char * takes only 1 byte so why is value at 11 address
printed( i mean 02) and not value at 10 address( 00)

similarly for the second output where u say p+1 goes from 10 to 11 due to
char type.when u do int* u say it dereferences 11 and 12. so then when char
* is done finally why does it print the one at 12 and not 11?
please correct my understanding if wrong


On Mon, Aug 15, 2011 at 10:44 AM, programming love <
love.for.programm...@gmail.com> wrote:

> The internal representation of array is this:
>
> suppose that the address starts from decimal number 10 and integer occupies
> 2 bytes
>
> 10- 0002 ( num 2 in hex)
> 12- 0003 ( num 3 in hex)
> 14- 0004 ( num 4 in hex)
>
> Now p points to address 10 and is type char. (Even after type casts) p+1
> will increment address by 1 byte (since it's char). p will now point to 11
> (int *) will say that when de-referenced 2 bytes should be extracted. So the
> 2 bytes extracted are 11, 12. Numbers in these bytes are 02 and 00
>
> 10- 00*02* ( num 2 in hex)
> 12- *00*03 ( num 3 in hex)
> 14- 0004 ( num 4 in hex)
>
> now (char *) says extract 1 byte for me. The extracted byte is 00. Hence 0
> is printed
>
> *Correct me if i am wrong.*
>
>
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