Consider this program:

        $a=5;
        print (++$a=0);
        print " $a\n";

        $a=5;
        print ((++$a)=0);
        print " $a\n";

        $a=5;
        print (++($a=0));
        print " $a\n";

You'd expect the first and second groups to print "0 0" and the third "1 1",
right?  Wrong.  3.0.1.5 on a Sparc, 4.0.1.1 on a Vax, and 4.0.1.4 on a Sparc
all generate this output:

        0 6
        0 0
        1 1

Is this really how it's supposed to work?

--
% =(one,hacker,two,Just,three,another,four,Perl4);$,=' ';print values% ,"\n";

Ah yes, the ++/= anomaly.  Here's a simpler form of it:
        $x = 0; (++ $x) = 5;    # sets $x to 5.
        $x = 0; ++ ($x = 5);    # sets $x to 6.
        $x = 0; ++ $x = 5;      # sets $x to 1!

The last line should act like the first line, but the evaluator ends
up assigning 5 to a useless location.

I don't understand Perl's evaluator enough to provide a safe fix.
--

| Ah yes, the ++/= anomaly.  Here's a simpler form of it:
|       $x = 0; (++ $x) = 5;    # sets $x to 5.
|       $x = 0; ++ ($x = 5);    # sets $x to 6.
|       $x = 0; ++ $x = 5;      # sets $x to 1!
|
| The last line should act like the first line, but the evaluator ends
| up assigning 5 to a useless location.

I too believe that the last line should act like the first line, and
be illegal, to parallel C's interpretation.

The result of a ++ or -- is not an lvalue.

for ("Juss","anotheq","Perk","hackeq") { print ++$_, " "; } print "\b,";
--
/=Randal L. Schwartz, Stonehenge Consulting Services (503)777-0095 ==========\
| on contract to Intel's iWarp project, Beaverton, Oregon, USA, Sol III      |

\=Cute Quote: "Intel: putting the 'backward' in 'backward compatible'..."====/

Hmm.  This is true in C, but Perl's grammar doesn't disallow it and
Perl doesn't complain about it, as it would if you said "3 = 4;".  It
just fails to work as you expect.

I don't see any reason why (++$x) shouldn't be an lvalue, but it
doesn't seem particularly useful.  Well, maybe double-post-increment
can be useful.
--

        Ah yes, the ++/= anomaly.  Here's a simpler form of it:
                $x = 0; (++ $x) = 5;    # sets $x to 5.
                $x = 0; ++ ($x = 5);    # sets $x to 6.
                $x = 0; ++ $x = 5;      # sets $x to 1!

        The last line should act like the first line, but the evaluator ends
        up assigning 5 to a useless location.

        I don't understand Perl's evaluator enough to provide a safe fix.

Ah yes, the 'try to get all in one statement anomaly'.  The safe
fix is that if it isn't obvious or the language could get
confused by the syntax, use separate statements!  Just because
the language will allow me to do something in one statement
doesn't mean I have to.  I don't know how many times I've shot
myself in the foot trying to get too cute with things like this.

Bill
--

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