Is this a known bug ?
Attached is small test prog that shows the problem.
char[] a = <whatever>;
char[] b = <whatever-else>;

if ( a == b )
{
// this code will be run as if the line above was
// if (a.length == b.length && a[0] == b[0])
// so this implies "pos" == "pzz" !!
}

"Mike Wynn" <Mike_member@pathlink.com> wrote in message news:add42g$4pa$1@digitaldaemon.com...
> Is this a known bug ?
> Attached is small test prog that shows the problem.
> char[] a = <whatever>;
> char[] b = <whatever-else>;
>
> if ( a == b )
> {
> // this code will be run as if the line above was
> // if (a.length == b.length && a[0] == b[0])
> // so this implies "pos" == "pzz" !!
> }

That's intended behavior. To get the behaviour you're looking for, use ===
(that's 3 =).

Walter wrote:

> "Mike Wynn" <Mike_member@pathlink.com> wrote in message news:add42g$4pa$1@digitaldaemon.com...
> > Is this a known bug ?
> > Attached is small test prog that shows the problem.
> > char[] a = <whatever>;
> > char[] b = <whatever-else>;
> >
> > if ( a == b )
> > {
> > // this code will be run as if the line above was
> > // if (a.length == b.length && a[0] == b[0])
> > // so this implies "pos" == "pzz" !!
> > }
>
> That's intended behavior. To get the behaviour you're looking for, use ===
> (that's 3 =).

????????
I thought that == was to compare the references!  Since "pos" and "pzz" must
be in different points in memory, they cannot have the same internal pointer,
and thus the test should fail, right?

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> > > char[] a = <whatever>;
> > > char[] b = <whatever-else>;
> > >
> > > if ( a == b )
> > > {
> > > // this code will be run as if the line above was
> > > // if (a.length == b.length && a[0] == b[0])
> > > // so this implies "pos" == "pzz" !!
> > > }
> >
> > That's intended behavior. To get the behaviour you're looking for, use
===
> > (that's 3 =).
> 
> ????????
> I thought that == was to compare the references!  Since "pos" and "pzz"
must
> be in different points in memory, they cannot have the same internal
pointer,
> and thus the test should fail, right?

The test

a[0] == b[0]

compares the values in the first element of the two arrays.

a[0] === b[0]

compares the references to the first element of the two arrays.

I think that's right...

Alix Pexton...

if (a == b) doesn't compare refferences

If that's correct, it's going to cause a nightmare in porting. And I know this has readly been fought out. On the otherhand I'd encourage programmers away from using pointers.


"Alix Pexton" <Alix@seven-point-star.co.uk> wrote in message news:01c20a8a$fec54160$fe247ad5@jpswm...
> > > > char[] a = <whatever>;
> > > > char[] b = <whatever-else>;
> > > >
> > > > if ( a == b )
> > > > {
> > > > // this code will be run as if the line above was
> > > > // if (a.length == b.length && a[0] == b[0])
> > > > // so this implies "pos" == "pzz" !!
> > > > }
> > >
> > > That's intended behavior. To get the behaviour you're looking for, use
> ===
> > > (that's 3 =).
> >
> > ????????
> > I thought that == was to compare the references!  Since "pos" and "pzz"
> must
> > be in different points in memory, they cannot have the same internal
> pointer,
> > and thus the test should fail, right?
>
> The test
>
> a[0] == b[0]
>
> compares the values in the first element of the two arrays.
>
> a[0] === b[0]
>
> compares the references to the first element of the two arrays.
>
> I think that's right...
>
> Alix Pexton...

"Russ Lewis" <spamhole-2001-07-16@deming-os.org> wrote in message news:3CFA9191.EE2B49FA@deming-os.org...
> ????????
> I thought that == was to compare the references!

It used to be. Now it's a deep compare, as everyone convinced me that was a better approach. === is for reference comparisons.

"Walter" <walter@digitalmars.com> wrote in message news:adeq4k$80d$1@digitaldaemon.com...

> "Russ Lewis" <spamhole-2001-07-16@deming-os.org> wrote in message news:3CFA9191.EE2B49FA@deming-os.org...
> > ????????
> > I thought that == was to compare the references!
>
> It used to be. Now it's a deep compare, as everyone convinced me that was
a
> better approach. === is for reference comparisons.

Then how does it turn out that "pos" == "psz" (see the first message in
the thread)?

"Pavel Minayev" <evilone@omen.ru> wrote in message news:adf7ca$o7m$1@digitaldaemon.com...
> "Walter" <walter@digitalmars.com> wrote in message news:adeq4k$80d$1@digitaldaemon.com...
>
> > "Russ Lewis" <spamhole-2001-07-16@deming-os.org> wrote in message news:3CFA9191.EE2B49FA@deming-os.org...
> > > ????????
> > > I thought that == was to compare the references!
> >
> > It used to be. Now it's a deep compare, as everyone convinced me that
was
> a
> > better approach. === is for reference comparisons.
>
> Then how does it turn out that "pos" == "psz" (see the first message in
> the thread)?

The array lengths and contents match. I guess I don't understand what the question is.

In article <adgof2$2de0$2@digitaldaemon.com>, Walter says...
>
>> Then how does it turn out that "pos" == "psz" (see the first message in
>> the thread)?
>
>The array lengths and contents match. I guess I don't understand what the question is.
>
the contents do not match "pos" is not the same contents as "psz"
if you see my first posting D only checks the first element in the
array if the lengths match.
surely it should check all then element for equality before
considering them equal.

Mike.

<mike.wynn@l8night.co.uk> wrote in message news:adh1r5$2ndr$1@digitaldaemon.com...
> In article <adgof2$2de0$2@digitaldaemon.com>, Walter says...
> >
> >> Then how does it turn out that "pos" == "psz" (see the first message in
> >> the thread)?
> >
> >The array lengths and contents match. I guess I don't understand what the question is.
> >
> the contents do not match "pos" is not the same contents as "psz"
> if you see my first posting D only checks the first element in the
> array if the lengths match.
> surely it should check all then element for equality before
> considering them equal.

Yes, it should check all the elements. If it doesn't, it's a bug.