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Home/ Questions/Q 6070249
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T09:54:43+00:00 2026-05-23T09:54:43+00:00

I am reading Damian Conway’s Perl Best Practices and found the following code snipet:

  • 0

I am reading Damian Conway’s “Perl Best Practices” and found the following code snipet:

$have_reconsidered{lc($name)}++;

I am trying to figure out what is going on here with the hash. I know ++ increments by one in a numeric context, but what does it do to a hash?

From perlop documentation:

undef is always treated as numeric,
and in particular is changed to 0
before incrementing (so that a
post-increment of an undef value will
return 0 rather than undef). The
auto-decrement operator is not
magical.

So in the example above, is the value for the key lc($name) being initialized to 0 and then incremented to 1 by ++?

In general, where could I find out more about the behaviors of ++, +=, etc…?

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  1. Editorial Team
    Editorial Team
    2026-05-23T09:54:43+00:00Added an answer on May 23, 2026 at 9:54 am

    %have_reconsidered is your hash. $name is a string. lc($name) returns the lowercased string. $hash{$key} will return the scalar value from hash %hash stored with key $key. so:

    // get scalar value from hash at key lc($name) and post-increment it
    $have_reconsidered(lc($name)}++;
    

    so, all you do is increment the value in a hash at a given index (namely lc($name))

    test case:

    #!/bin/env perl
    my %hash = ( 'a' => '2' );
    my $name = 'A';
    print $hash{lc($name)}++; // prints 2 (incremented after statement)
    print $hash{lc($name)};   // prints 3
    print ++$hash{lc($name)}; // prints 4 (incremented before statement)
    
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