Question about ternary operator

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

Hello Monks, Can someone explain why the first ternary statement produces something completely different? The two ternary statements should be quivalent to the if else block that follows ... or not? I would expect this output:

rc_OK (=1)
rc_OK (=1)
====
rc_OK (=1)
[download]

But instead I get this output:

check_group GRP_ADL001 0
rc_OK (=1)
====
rc_OK (=1)
[download]

code snippet

use strict;
use Data::Dumper;

my $checked;
my $GroupDN="GRP_ADL001";

$checked->{Group}{$GroupDN}=1;
#print Dumper($checked);

my $rc_OK;

exists($checked->{Group}{$GroupDN}) ? $rc_OK=1 : $rc_OK=check_group($G
+roupDN,0); # this does NOT work
$rc_OK=exists($checked->{Group}{$GroupDN}) ? 1 : check_group($GroupDN,
+0); # this does work

print "rc_OK (=$rc_OK)\n";

print "====\n";

if(exists($checked->{Group}{$GroupDN})){
    $rc_OK=1
}
else{
    $rc_OK=check_group($GroupDN,0);
}        
print "rc_OK (=$rc_OK)\n";

sub check_group{
    my($group,$rc)=@_;
    print "check_group $group $rc\n";
    return($rc);
}
[download]

Comment on Question about ternary operator Select or Download Code

Replies are listed 'Best First'.
Re: Question about ternary operator by moritz (Cardinal) on Jul 01, 2009 at 08:32 UTC
The problem is that the assignment has a looser precedence than the ternary op. See also: perlop, Confused by Perl ternary operator, ternary operator. Why does the Perl conditional operator not do what I expect? and many other threads.	[reply]
Re^2: Question about ternary operator by merlinX (Novice) on Jul 01, 2009 at 08:42 UTC
Thx, operator precedence is sometimes a PITA, so this is what the first ternary statement should look like then `exists($checked->{Group}{$GroupDN}) ? ($rc_OK=1) : ($rc_OK=check_group +($GroupDN,0));` [download]	[reply] [d/l]
Re^3: Question about ternary operator by moritz (Cardinal) on Jul 01, 2009 at 08:52 UTC
No, it should look exactly like the first statement. The ternary operator is very good if you use its value, but if you use it in void context it's just a kludge for an `if` statement. So `# bad: ternary in void context: exists($checked->{Group}{$GroupDN}) ? ($rc_OK=1) : ($rc_OK=check_group +($GroupDN,0)); # good: 'if' in void context: if (exists($checked->{Group}{$GroupDN})) { $rc_OK = 1 } else { $rc_OK=check_group($GroupDN,0) } # also good: ternary in scalar context: $rc_OK = exists($checked->{Group}{$GroupDN}) ? 1 : check_group($Group +DN,0)` [download] So you have two good, idiomatic ways to express something, and complain that the third is a PITA?	[reply] [d/l] [select]
Re^3: Question about ternary operator by ikegami (Patriarch) on Jul 01, 2009 at 18:31 UTC
Thx, operator precedence is sometimes a PITA If the precedence was reversed, the intended use would not work. `$y = $c ? $x1 : $x2;` [download] would mean `($y = $c) ? $x1 : $x2;` [download] By the way, it's called the "conditional operator". Perl has many ternary operators.	[reply] [d/l] [select]
Re^4: Question about ternary operator by Anonymous Monk on Jul 01, 2009 at 18:38 UTC
Re^5: Question about ternary operator by ikegami (Patriarch) on Jul 01, 2009 at 19:02 UTC
Re^2: Question about ternary operator by rovf (Priest) on Jul 01, 2009 at 11:50 UTC
The problem is that the assignment has a looser precedence than the ternary op. Shouldn't then the OP's program had reported a syntax error? -- Ronald Fischer <ynnor@mm.st>	[reply] [d/l]
Re^3: Question about ternary operator by moritz (Cardinal) on Jul 01, 2009 at 11:54 UTC
No: `$ perl -MO=Deparse,-p -e '1 ? $a = 3 : $b = 4' (($a = 3) = 4); -e syntax OK` [download] You see it is parsed, not just the way you'd naively expect it.	[reply] [d/l]