perldata doesn't tell what happens on list assignment, with less/more elements #16289
Comments
From [email protected]This is a bug report for perl from calestyo@scientia.net, perldata explains most things related to lists, arrays, etc. and However, I feel it misses one crucial point, namely what happens @more=(1, 2, 3, 4); and with that each of the following then the cases: Seems the result is, that if there is - as above - no array or hash Cheers, Flags: Site configuration information for perl 5.26.1: Configured by Debian Project at Tue Nov 28 17:44:14 UTC 2017. Summary of my perl5 (revision 5 version 26 subversion 1) configuration: Locally applied patches: @INC for perl 5.26.1: Environment for perl 5.26.1: |
From @jkeenanOn Thu, 07 Dec 2017 04:00:45 GMT, calestyo@scientia.net wrote:
Please review the patch attached. Thank you very much. -- |
From @jkeenan132538-0001-Clarify-different-cases-of-assignment-to-list-of-sca.patchFrom e01ff6d79771f0c623f6be7a477289c8edd26d35 Mon Sep 17 00:00:00 2001
From: James E Keenan <[email protected]>
Date: Thu, 7 Dec 2017 16:09:54 -0500
Subject: [PATCH] Clarify different cases of assignment to list of scalars.
Prepared in response to RT #132538.
---
pod/perldata.pod | 50 ++++++++++++++++++++++++++++++++++++++++++++++++++
1 file changed, 50 insertions(+)
diff --git a/pod/perldata.pod b/pod/perldata.pod
index 9d714f2..b093e3a 100644
--- a/pod/perldata.pod
+++ b/pod/perldata.pod
@@ -778,6 +778,56 @@ As of Perl 5.22, you can also use C<(undef)x2> instead of C<undef, undef>.
(You can also do C<($x) x 2>, which is less useful, because it assigns to
the same variable twice, clobbering the first value assigned.)
+Note that assignment of one list of scalars to another list of scalars is
+B<not> the same as assigning that first list to an array.
+
+When you assign a list of scalars to an array, all previous values in that
+array are wiped out and number of elements in the array will now be equal to
+the number of elements in the right-hand list -- the list from which
+assignment was made. The array will automatically resize itself to precisely
+accommodate each element in the right-hand list.
+
+ use warnings;
+ my (@xyz, $x, $y, $z);
+
+ @xyz = (1, 2, 3);
+ print "@xyz\n"; # 1 2 3
+
+ @xyz = ('al', 'be', 'ga', 'de');
+ print "@xyz\n"; # al be ga de
+
+ @xyz = (101, 102);
+ print "@xyz\n"; # 101 102
+
+When, however, you assign a list of scalars to another list of scalars, the
+results differ according to whether the left-hand list -- the list being
+assigned to -- has the same, more or fewer elements than the right-hand list.
+
+ ($x, $y, $z) = (1, 2, 3);
+ print "$x $y $z\n"; # 1 2 3
+
+ ($x, $y, $z) = ('al', 'be', 'ga', 'de');
+ print "$x $y $z\n"; # al be ga
+
+ ($x, $y, $z) = (101, 102);
+ print "$x $y $z\n"; # 101 102
+ # Use of uninitialized value $z in concatenation (.)
+ # or string at [program] line [line number].
+
+If the number of scalars in the left-hand list is less than that in the
+right-hand list, the "extra" scalars in the right-hand list will simply not be
+assigned.
+
+If the number of scalars in the left-hand list is greater than than in the
+left-hand list, the "missing" scalars will become undefined.
+
+ ($x, $y, $z) = (101, 102);
+ for my $el ($x, $y, $z) {
+ (defined $el) ? print "$el " : print "<undef>";
+ }
+ print "\n";
+ # 101 102 <undef>
+
List assignment in scalar context returns the number of elements
produced by the expression on the right side of the assignment:
--
2.7.4
|
|
The RT System itself - Status changed from 'new' to 'open' |
From @cpansproutOn Thu, 07 Dec 2017 13:18:52 -0800, jkeenan wrote:
I find this sentence confusing. I think the rest of your patch clarifies things sufficiently that you can just leave out this sentence.
*the* number
s/than tha\Kn/t/ -- Father Chrysostomos |
From @jkeenanOn Fri, 08 Dec 2017 00:55:01 GMT, sprout wrote:
Pushed to blead with those corrections in commit 436908e. Marking ticket Resolved. Thank you very much. -- |
|
@jkeenan - Status changed from 'open' to 'resolved' |
From [email protected]Hi. Thanks for the patch :) My comments below: +Note that assignment of one list of scalars to another list of scalars is => I also think this is confusing... +When you assign a list of scalars to an array, all previous values in that => I would add here (just for convenience): + use warnings; + my (@xyz, $x, $y, $z); => could also fit in 3 lines, but I'm fine with it :) + => "... or an array" , the => one might add here, that 'de' is of course still evaluated, but just discarded + => there are two times left-hand list... the 2nd should be right hand list + Thanks :-) |
Migrated from rt.perl.org#132538 (status was 'resolved')
Searchable as RT132538$
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