whats wrong reverse %x = reverse %h?

borisz has asked for the wisdom of the Perl Monks concerning the following question:

Hi, I typed:

perl -e '%h = ( q => 2, w => 2 ); print reverse %h = reverse %h'
[download]

and wonder why I get three values back 'w22' I expect 'q2' or 'w2' at perls option. More testing shows even more wired results. In the example below, I expect %x reversed in %h but it is not. There was not even a 'a' => 2 pair in the original %h.

perl -MData::Dumper -le '%h = (1 => 2, a=>"b", c => 2); %h = reverse (
+ %x = reverse %h); print Dumper({x => \%x, h =>\%h});'
__OUTPUT__
$VAR1 = {
          'x' => {
                   'b' => 'a',
                   '2' => 'c'
                 },
          'h' => {
                   'a' => 2,
                   'c' => 2
                 }
        };
[download]

Perl 5.8.4 cofuses me even more

perl5.8.4 -MData::Dumper -le '%h = (1 => 2, a=>"b", c => 2); %h = reve
+rse ( %x = reverse %h); print Dumper({x => \%x, h =>\%h});'
__OUTPUT__
$VAR1 = {
          'x' => {
                   'b' => 'a',
                   '2' => 'c'
                 },
          'h' => {
                   '' => 2
                 }
        };
[download]

can you explain this behavior please?
UPDATE: I really expect to lose the dupes. My point is that I expect that %h is a reverse of %x. But it is not for me, tested on
Mac OSX and Linux. So I really wonder that you get different output.

Boris

Comment on whats wrong reverse %x = reverse %h? Select or Download Code

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Re: whats wrong reverse %x = reverse %h? by blokhead (Monsignor) on Sep 28, 2004 at 16:52 UTC
The problem is with using the expression `(%x = reverse %h)` in list context. Notice that when I try to use it to make an anonymous array: `use Data::Dumper; my %h = (1 => 2, a => "b", c => 2); print Dumper [ my %x = reverse %h ];` [download] I get a `Bizarre copy of ARRAY in anonlist` (perl v5.8.3). Same if I try to do `print Dumper { my %x = reverse %h };` [download] I have no idea what `(my %x = reverse %h)` evaluates to in list context, but whatever it is, Perl doesn't like it. The fact it doesn't die with the `Bizarre copy` error when using that expression as function arguments might be a bug... Anyway, back to your original question, if you split the two reverses into two statements like `%h = reverse do { my %q = reverse %h; %q };` [download] you'll get what you expect. blokhead	[reply] [d/l] [select]
Re: whats wrong reverse %x = reverse %h? by waswas-fng (Curate) on Sep 28, 2004 at 16:48 UTC
Deparsed output: `perl -MO=Deparse -MData::Dumper -le '%h = (1 => 2, a=>"b", c => 2); %h + = reverse ( %x = reverse %h); print Dumper({x => \%x, h =>\%h});' BEGIN { $/ = "\n"; $\ = "\n"; } use Data::Dumper; (%h) = (1, 2, 'a', 'b', 'c', 2); (%h) = reverse((%x) = reverse(%h)); print Dumper({'x', \%x, 'h', \%h}); -e syntax OK` [download] as stated above your reverse/reverse of the hash is causing you to lose dup keys... -Waswas	[reply] [d/l]
Re^2: whats wrong reverse %x = reverse %h? by borisz (Canon) on Sep 28, 2004 at 16:51 UTC
Dont get me wrong, I really expect to lose the dup keys, but my output lose much more then the dupes. Boris	[reply]
Re: whats wrong reverse %x = reverse %h? by Zaxo (Archbishop) on Sep 28, 2004 at 16:40 UTC
When you reverse the list you get from a hash, keys and values are exchanged. Making a hash from the reversed list, if two values are the same, only one of the previous keys will survive as a value. Added - I get 'w2' running your first example, but get the same as you with the second. There may be some optimization with aliases going wrong. I don't see that the construction is supposed to be undefined. After Compline, Zaxo	[reply]
Re^2: whats wrong reverse %x = reverse %h? by borisz (Canon) on Sep 28, 2004 at 16:47 UTC
Thats what I expect, but none of the examples show this behavior. the expected results from the first is for %h `{ 2 => 1, b=> 'a' } or { 2 => 'c', b=> 'a' }` [download] but the result is `{ a => 2, b => 2}` [download] even more unexpeted results from perl 5.8.4 where I get `{ '' => '2' }` [download] Boris	[reply] [d/l] [select]
Re: whats wrong reverse %x = reverse %h? by melora (Scribe) on Sep 28, 2004 at 21:19 UTC
I did a little playing with this, to try to learn something. `%h = (1 => 2, a=>"b", c => 2); %h = reverse (%x = reverse %h); foreach my $n (keys %h) { print "$n $h{$n}\n"; }` [download] Yeah, I know. Anyway, here's what I get: `2 1 2 a b` [download] As if a duplicate element were eliminated by making the key null. But if I do `%h = (1 => 2, a=>"b", c => 2); %x = reverse %h; %h = reverse %x; foreach my $n (keys %h) { print "$n $h{$n}\n"; }` [download] I get `1 2 a b` [download] which is what I'd expect. Now, here's yet another thing: if I create another duplicate, d => 2 (at the end of the list), I still get the same end result; that duplicate is eliminated. In fact, in the first bit of code, I still get three lines of output. If there are duplicates, it doesn't seem to be guaranteed which one is eliminated by the reverse. I just looked at reverse in my Camel book and it doesn't mention anything about duplicates, so I suspect the order can't be relied upon. Typical for a hash.	[reply] [d/l] [select]
Re: whats wrong reverse %x = reverse %h? by jZed (Prior) on Sep 28, 2004 at 16:44 UTC
I get 'w2' (no second 2) using perl 5.8.3. on winXP.	[reply]
Re^2: whats wrong reverse %x = reverse %h? by hardburn (Abbot) on Sep 28, 2004 at 16:47 UTC
Same for me on perl, v5.8.4 built for i686-linux. "There is no shame in being self-taught, only in not trying to learn in the first place." -- Atrus, Myst: The Book of D'ni.	[reply]
Re: whats wrong reverse %x = reverse %h? by Keystroke (Scribe) on Sep 29, 2004 at 14:55 UTC
Maybe it has something to do with print being on the same line? Weird. `% perl -e '%h = (q => 1, w => 2, x => 0, y => 2);\ print join ",", reverse %h = reverse %h;' w,2,x,0,,2,q,21 % perl -e '%h = (q => 1, w => 2, x => 0, y => 2);\ %h = reverse %h; print join ",", reverse %h;' w,2,x,0,q,21` [download]	[reply] [d/l]


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