Re^2: Count number of elements in HoH and sum them for other keys

thank you so much. I was a bit confused at the beginning: I thought that you had to specify a starting value for $hoh{$c1}{$c2}{count}. By the way, and I know I'm going a bit astray of the initial question, but what if I wanted to start that count at 5? Would specifying

$hoh{$c1}{count}=5;
[download]

work if I specified it before incrementing in your while loop? Thanks!

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Re^3: Count number of elements in HoH and sum them for other keys by smls (Friar) on Jun 03, 2014 at 13:21 UTC
I thought that you had to specify a starting value for $hoh{$c1}{$c2}{count}. When you dereference or modify a non-existing array or hash element, it will automatically "spring to life", including all the necessary intermediate hashes/arrays. For example: `my %test; $test{a}[2]{b} = 'Hello'; # %test now contains: # ( a => [ undef, # undef, # { b => "Hello" } ] )` [download] It's called autovivification, and it's one of the nice features that make Perl special... :) See Wikipedia and perlreftut for more info. In addition, the ++ (auto-increment) operator silently treats `undef` as `0`. So you don't need to specify an initial value. what if I wanted to start that count at 5? One solution would be to create the hash first, and then use another loop to add 5 to each counter. Alternatively, you can do a check inside the loop (before incrementing!) to see if the counter has already been incremented previously, and if not, initialize it with the number 5: `if (!$hoh{$c1}{count}) { $hoh{$c1}{count} = 5; } # verbose form` [download] `$hoh{$c1}{count} \|\|= 5; # shortcut` [download] (See C style Logical Or and Assignment Operators.)	[reply] [d/l] [select]

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Re^3: Count number of elements in HoH and sum them for other keys
by smls (Friar) on Jun 03, 2014 at 13:21 UTC

I thought that you had to specify a starting value for $hoh{$c1}{$c2}{count}.

When you dereference or modify a non-existing array or hash element, it will automatically "spring to life", including all the necessary intermediate hashes/arrays. For example:

my %test;
$test{a}[2]{b} = 'Hello';

# %test now contains:
#   ( a => [ undef,
#            undef,
#            { b => "Hello" } ] )
[download]

It's called autovivification, and it's one of the nice features that make Perl special... :) See Wikipedia and perlreftut for more info.

In addition, the ++ (auto-increment) operator silently treats undef as 0. So you don't need to specify an initial value.

what if I wanted to start that count at 5?

One solution would be to create the hash first, and then use another loop to add 5 to each counter.

Alternatively, you can do a check inside the loop (before incrementing!) to see if the counter has already been incremented previously, and if not, initialize it with the number 5:

if (!$hoh{$c1}{count}) { $hoh{$c1}{count} = 5; }   # verbose form
[download]

$hoh{$c1}{count} ||= 5;   # shortcut
[download]

(See C style Logical Or and Assignment Operators.)

[reply]
[d/l]
[select]