Assign a variable and check existence of hash key


With the little that I know about Perl, I was surprised that this syntax doesn't work:

if (exists $wid = $dict{$w}) { print "$wid:$c\n"; }

It seems I can't make the assignment $wid = $dict{$w} and check if it exists at the same time:


exists argument is not a HASH or ARRAY element or a subroutine at createWordIndex.pl line 31.


Is there a way to do this kind of assignment? Or do I just need to check existence first and then make the assignment if the condition is true?


Are you trying to figure out if there's a value there before you print it? Then you'll have to do it this way:

if ( exists($dict{$w}) ) { my $wid = $dict{$w}; print "$wid: $c\n"; }

If you know that the values stored will always be defined, you can do

if ( defined( my $wid = $dict{$w} ) ) { print "$wid: $c\n"; }


Alternatively (TIMTOWTDI, after all), you can use the smart match operator (for Perl version >= 5.10):

if($w ~~ (keys %dict)) { my $wid = $dict{$w}; print "$wid: $c\n"; }

Or, alternatively:

if(grep{$w eq $_}(keys %dict)) { my $wid = $dict{$w}; print "$wid: $c\n"; }


You just answered your question in your question.


"Or do I just need to check existence first and then make the assignment if the condition is true?"


the <strong>exists</strong> function in your code takes in the scalar $wid. And is clearly stated in your error that the exists function needs a hash or array element


exists argument is not a HASH or ARRAY element or a subroutine at createWordIndex.pl line 31.


so yes. you need to check first if it exists.

friendly advice: try all the things/solutions that you can think of before asking because you might already know the answer to your question. :)


you can do it, just skip the exists. if the assignment is made (because there is a $dict{$w}) the side-effect is true.

my %dict; $dict{one} = 1; $dict{two} = 2; $dict{three} = 3; if ($wid = $dict{four}) { print "$wid\n"; } else { print "NO!\n"; }


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