Re: Checking if hash value exists without implicitly defining it

in reply to Checking if hash value exists without implicitly defining it

One option is to short circuit the autovivification:

use strict;
use warnings;

my %Hash;
$Hash{44.52}{param1}{key1} = [ 101.01, 'val1' ];
$Hash{95.01}{param2}{key2} = [ 101.02, 'val2' ];

my $neededVal = 80.0;

if (    exists $Hash{$neededVal}
    and exists $Hash{$neededVal}{param1}
    and exists $Hash{$neededVal}{param1}{key1} )
{
    print "DEFINED";
}
else {
    for ( keys %Hash ) {
        print "$_\n";
    }
}
[download]

Output:

95.01
44.52
[download]

Update I: Added the needed defined, in case of an earlier (for example) $Hash{$neededVal}{param1}{key1} = 0.

Update II: Replaced defined with exists. Thank you, choroba.

Comment on Re: Checking if hash value exists without implicitly defining it Select or Download Code
Re^2: Checking if hash value exists without implicitly defining it by Anonymous Monk on Feb 14, 2013 at 16:40 UTC
Huh. I thought you needed to use exists to avoid AVing the hash key you're reading, but that code works. This should simplify some of my code.	[reply]
Re^3: Checking if hash value exists without implicitly defining it by choroba (Prior) on Feb 14, 2013 at 17:26 UTC
Well, it only breaks for the false values like 0 or "" or `undef`. لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ	[reply] [d/l]
Re^3: Checking if hash value exists without implicitly defining it by Kenosis (Deacon) on Feb 14, 2013 at 17:42 UTC
My apologies, as it should be written as: `... if ( exists $Hash{$neededVal} and exists $Hash{$neededVal}{param1} and exists $Hash{$neededVal}{param1}{key1} ) { ...` [download] Will correct the original... Update: Replaced `defined` with `exists`. Thank you, choroba.	[reply] [d/l] [select]
Re^4: Checking if hash value exists without implicitly defining it by choroba (Prior) on Feb 14, 2013 at 17:44 UTC
What is wrong with plain old exists? لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ	[reply]
Re^5: Checking if hash value exists without implicitly defining it by Kenosis (Deacon) on Feb 14, 2013 at 17:52 UTC
Re^2: Checking if hash value exists without implicitly defining it by Anonymous Monk on Feb 15, 2013 at 10:04 UTC
You don't even need the `exists` (assuming you don't have a weird freeform data structure): `if (Hash{$neededVal} and $Hash{$neededVal}{param1} and $Hash{$neededVal}{param1}{key1} > 4) {` [download]	[reply] [d/l] [select]
Re^3: Checking if hash value exists without implicitly defining it by Kenosis (Deacon) on Feb 15, 2013 at 15:47 UTC
Yes, you're correct! Originally omitted `exists` when not testing for the final value--then overgeneralized. However, testing `$Hash{$neededVal}{param1}{key1}` would fail when `$Hash{$neededVal}{param1}{key1} == 0`, but expliciting testing the value, as you've shown, would remedy that.	[reply] [d/l] [select]


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Re: Checking if hash value exists without implicitly defining it

How many continents have you visited?