«

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

Yes you are right.

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

wrote:
> Is *ptr++ equivalent to *(ptr++) ?

yes it is correct....
one more

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

Richard Heathfield <> wrote:
> 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 "associativity" 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 LID | sed s/[[:upper:]]//g`

wrote:

>
>
> 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.

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


<> wrote in message
news: ps.com...
> Is *ptr++ equivalent to *(ptr++) ?
>

Chris Dollin wrote:
> wrote:
>
> >
> >
> > 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/

Vijay Kumar R. Zanvar wrote:

>
>
> Chris Dollin wrote:
>> wrote:
>>
>> >
>> >
>> > 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.

Chris Dollin wrote:
> Vijay Kumar R. Zanvar wrote:
>
> >
> >
> > Chris Dollin wrote:
> >> wrote:
> >>
> >> >
> >> >
> >> > 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.