is *ptr++ equivalent to *(ptr++)

[junky_fellow]

Is *ptr++ equivalent to *(ptr++) ?

[Rajan]

Yes you are right.

[Richard Heathfield]

ju**********@ya hoo.co.in wrote:

Is *ptr++ equivalent to *(ptr++) ?

Yes. In each case, the associativity is right-to-left and, in each case, the
value seen by * is the old value of ptr, not the new value that it will
have when ++ has completed its work.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
mail: rjh at above domain

[ranjeet.gupta]

ju**********@ya hoo.co.in wrote:

Is *ptr++ equivalent to *(ptr++) ?

yes it is correct....
one more

++*ptr == (*ptr)++

[S.Tobias]

Richard Heathfield <in*****@addres s.co.uk.invalid > wrote:

ju**********@ya hoo.co.in wrote:

Is *ptr++ equivalent to *(ptr++) ?
Yes. In each case, the associativity is right-to-left and, in each case, the

I don't think "associativ ity" is the right word here, as they are
unary operators. "Precedence " is more appropriate, I think.
value seen by * is the old value of ptr, not the new value that it will
have when ++ has completed its work.

There's a rule of thumb for unary operators in C, first you read ones
to the right, then ones to the left (unless, of cource, paretheses
override this).

A couple of (non-trivial) examples:
*p++
++p[i];
*p[i]
&a[i]
++p->m
(type*)p->m
*p()

(But I haven't really made sure that it always works; and `sizeof' is
special.)

--
Stan Tobias
mailx `echo si***@FamOuS.Be dBuG.pAlS.INVALID | sed s/[[:upper:]]//g`

[Chris Dollin]

ra***********@g mail.com wrote:

ju**********@ya hoo.co.in wrote:

Is *ptr++ equivalent to *(ptr++) ?

yes it is correct....
one more

++*ptr == (*ptr)++

I hardly think so.

--
Chris "electric hedgehog" Dollin
It's called *extreme* programming, not *stupid* programming.

[Eric]

Yes. See the details here:
http://msdn.microsoft.com/library/de....operators.asp
(google with "operator precedence" and "left to right" and "right to left"
for more lists)

Like you can see in the list, the post-increment operator will be executed
first, if there are no brackets.

The associativity for the indirection operator is right to left and for the
post-increment operator left to right. But for this example this is not of
any importance.

Eric
<ju**********@y ahoo.co.in> wrote in message
news:11******** ************@g4 3g2000cwa.googl egroups.com...

Is *ptr++ equivalent to *(ptr++) ?

[Vijay Kumar R. Zanvar]

Chris Dollin wrote:

ra***********@g mail.com wrote:

ju**********@ya hoo.co.in wrote:

Is *ptr++ equivalent to *(ptr++) ?

yes it is correct....
one more

++*ptr == (*ptr)++

I hardly think so.

The two constructs are equivalent in the following scenarios:

int *ptr = ...;
int i;

(void)++*ptr;
(void)(*ptr)++;

However, their behaviour is not the same when used like this:

i = (*ptr)++;
i = ++*ptr;

--
Vijay Kumar R. Zanvar
Home Page - http://geocities.com/vijoeyz/

[Chris Dollin]

Vijay Kumar R. Zanvar wrote:

Chris Dollin wrote:

ra***********@g mail.com wrote:

>
>
> ju**********@ya hoo.co.in wrote:
>> Is *ptr++ equivalent to *(ptr++) ?
>
> yes it is correct....
> one more
>
> ++*ptr == (*ptr)++

I hardly think so.

The two constructs are equivalent in the following scenarios:

int *ptr = ...;
int i;

(void)++*ptr;
(void)(*ptr)++;

However, their behaviour is not the same when used like this:

i = (*ptr)++;
i = ++*ptr;

Hence, they are not equivalent; you can't just use one in place
of the other.

--
Chris "electric hedgehog" Dollin
It's called *extreme* programming, not *stupid* programming.

[Vijay Kumar R. Zanvar]

Chris Dollin wrote:

Vijay Kumar R. Zanvar wrote:

Chris Dollin wrote:

ra***********@g mail.com wrote:

>
>
> ju**********@ya hoo.co.in wrote:
>> Is *ptr++ equivalent to *(ptr++) ?

[..]
Hence, they are not equivalent; you can't just use one in place
of the other.

Yes, you're quite correct. But for a learner, he/she must know the
difference. Practically, these two construct should not be
interchanged.
--
Chris "electric hedgehog" Dollin
It's called *extreme* programming, not *stupid* programming.