strange output from conditional operator

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

Hello, fellow monks;

I've gotten some strange results for some code I'm writing. I've made a short test case which I hope somebody can explain the results of to me.

use strict;
use warnings;

my @arr = ('one', 'apple', 'two', 'apple', 'three');


my $var;
my $matches = 0;
foreach my $el (@arr) {
    (defined($var) ? $var .= 'y' : $var .= 'n') if $el eq 'apple';
    $matches++ if $el eq 'apple';
}
print "$var\n";
print "$matches\n";

$var = 'x';
$matches = 0;
foreach my $el (@arr) {
    (defined($var) ? $var .= 'y' : $var .= 'n') if $el eq 'apple';
    $matches++ if $el eq 'apple';
}
print "$var\n";
print "$matches\n";
[download]

Output:

nyn
2
xynyn
2
[download]

Nevermind rewriting the code bringing the assignment/appending operator outside of the comparison. I would like to know why this specific code prints such strange results (I would expect 'ny' for the former and 'xyy' for the latter fragment).

I understand '=' and '.=' can return something, but I don't understand how it gets assigned back to $var or why the non-matching part of the conditional gets executed.

What am I missing here?

Comment on strange output from conditional operator Select or Download Code

Replies are listed 'Best First'.
Re: strange output from conditional operator by moritz (Cardinal) on May 15, 2009 at 11:43 UTC
You were bitten by the fact that the ternary operator `? :` has a tighter precedence than the assignment operator. Instead, try this: `$var .= (defined($var) ? 'y' : 'n';` [download] See also Confused by Perl ternary operator, perlop.	[reply] [d/l] [select]
Re^2: strange output from conditional operator by december (Pilgrim) on May 15, 2009 at 12:23 UTC
Yes that works – I knew it does – but I was wondering why doing it this way caused such strange output... I've given thought to operator precedence, but I figured things wouldn't align up and I'd get a warning or error. So how would this get parsed? Something like the following I presume: `(defined($var) ? $var) .= ('y' : $var .= 'n') if $el eq 'apple';` Thanks for the explanation, operator precedence makes a lot more sense now. I'm just surprised things lined up well enough not to cause any warnings or syntax errors.	[reply] [d/l]
Re^3: strange output from conditional operator by moritz (Cardinal) on May 15, 2009 at 12:31 UTC
The thread I've linked to above has an example both for how something very similar gets parsed, and how to find out how it's parsed. Did you not read it?	[reply]
Re^3: strange output from conditional operator by JadeNB (Chaplain) on May 15, 2009 at 21:53 UTC
`(defined($var) ? $var) .= ('y' : $var .= 'n') if $el eq 'apple';` No, because there can be no `?` without an accompanying `:`. Thus, the ‘middle’ assignment is parsed as you like (since the tokeniser hasn't seen a `:` yet, hence knows that it's still inside the conditional), but the ‘last’ one isn't (since the tokeniser believes it's done with the conditional after finding the second occurrence of `$var`): `(defined($var) ? $var .= 'y' : $var) .= 'n'` (Of course, the statement about every `?` requiring an accompanying `:` is a lie, as regular expressions show; but you know what I mean. :-) )	[reply] [d/l] [select]


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