«

"vjay" <> wrote in message
news: lkaboutprogramming.com...
> Check the code below guys
> void main()
> {
> char a[5] = {1,2,3,4,5};
> char *ptr;
> ptr = (char*)(&a + 1);
> printf("%d",*(ptr-1));
> }
>
> The output of the program is 5

Thanks, your post and it's replies has clarified something that I hadn't
fully grasped; in int a[5], then:

&a is not the same thing as a, when considering Type.

Apparently a is the same as &a[0] (pointer to int), &a means a pointer to
the entire array (pointer to array of 5 ints). The values however are the
same. (Although why &a doesn't decompose to &(&a[0]) is another mystery;
maybe it only decomposes when there isn't an & in front)

> void main()

With quotes, this gives 5 times as many Google hits as "int main(void)".
Looks like a lot of people can't get it right.

Bart

Bart C wrote:
>
> "vjay" <> wrote in message
> news: lkaboutprogramming.com...
> > Check the code below guys
> > void main()
> > {
> > char a[5] = {1,2,3,4,5};
> > char *ptr;
> > ptr = (char*)(&a + 1);
> > printf("%d",*(ptr-1));
> > }
> >
> > The output of the program is 5
>
> Thanks, your post and it's replies
> has clarified something that I hadn't
> fully grasped; in int a[5], then:
>
> &a is not the same thing as a, when considering Type.
>
> Apparently a is the same as &a[0] (pointer to int),
> &a means a pointer to
> the entire array (pointer to array of 5 ints).
> The values however are the same.
> (Although why &a doesn't decompose to &(&a[0]) is another mystery;
> maybe it only decomposes when there isn't an & in front)

That's very close to being correct.

N869
6.3.2 Other operands
6.3.2.1 Lvalues and function designators

[#3] Except when it is the operand of the sizeof operator or
the unary & operator, or is a string literal used to
initialize an array, an expression that has type ``array of
type'' is converted to an expression with type ``pointer to
type'' that points to the initial element of the array
object and is not an lvalue.

--
pete

Thank you all guys now i got it.

--
Message posted using http://www.talkaboutprogramming.com/group/comp.lang.c/
More information at http://www.talkaboutprogramming.com/faq.html

"Tomás Ó hÉilidhe" <t...@lavabit.com> wrote:
> "vjay" <vijayanand...@gmail.com> wrote
> > char a[5] = {1,2,3,4,5};
>
> Unusual to store numbers in a char, ...

What else does one store in a char?

--
Peter

vjay wrote:
>
> Check the code below guys
> void main()
> {
> char a[5] = {1,2,3,4,5};
> char *ptr;
> ptr = (char*)(&a + 1);
> printf("%d",*(ptr-1));
> }
>
> The output of the program is 5
>
> But what i expected was 1.When i replace &a with a in the expression ptr =
> (char*)(a + 1); i get output as 1.what is the difference between (&a + 1)
> and (a + 1) in the above mentioned expression.Help guys its confusing.

Consider:

What does it mean when you say "&something + 1"?

What is the value of "sizeof a"?

If there were a "typeof" operator, what would be "typeof a"?

--
+-------------------------+--------------------+-----------------------+
| Kenneth J. Brody | www.hvcomputer.com | #include |
| kenbrody/at\spamcop.net | www.fptech.com | <std_disclaimer.h> |
+-------------------------+--------------------+-----------------------+
Don't e-mail me at: <private.php?do=newpm&u=>

"Bart C" <> writes:
[...]
> Apparently a is the same as &a[0] (pointer to int), &a means a pointer to
> the entire array (pointer to array of 5 ints). The values however are the
> same. (Although why &a doesn't decompose to &(&a[0]) is another mystery;
> maybe it only decomposes when there isn't an & in front)
[...]

The answers to these and a number of similar questions can be found in
section 6 of the comp.lang.c FAQ, <http://www.c-faq.com>.

--
Keith Thompson (The_Other_Keith) <kst->
Looking for software development work in the San Diego area.
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"