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Home/ Questions/Q 258215
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T22:09:43+00:00 2026-05-11T22:09:43+00:00

I sometimes access a hash like this: if(exists $ids{$name}){ $id = $ids{$name}; } Is

  • 0

I sometimes access a hash like this:

if(exists $ids{$name}){
    $id = $ids{$name};
}

Is that good practice? I’m a bit concerned that it contains two lookups where really one should be done. Is there a better way to check the existence and assign the value?

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  1. Editorial Team
    Editorial Team
    2026-05-11T22:09:43+00:00Added an answer on May 11, 2026 at 10:09 pm

    By checking with exists, you prevent autovivification. See Autovivification : What is it and why do I care?.

    UPDATE: As trendels points out below, autovivification does not come into play in the example you posted. I am assuming that the actual code involves multi-level hashes.

    Here is an illustration:

    #!/usr/bin/perl
    
    use strict;
    use warnings;
    
    use Data::Dumper;
    
    my (%hash, $x);
    
    if ( exists $hash{test}->{vivify} ) {
        $x = $hash{test}->{vivify}->{now};
    }
    
    print Dumper \%hash;
    
    $x = $hash{test}->{vivify}->{now};
    
    print Dumper \%hash;
    
    __END__
    
    
    C:\Temp> t
    $VAR1 = {
        'test' => {}
    };
    $VAR1 = {
        'test' => {
            'vivify' => {}
        }
    };
    
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