autovivification

Suppose you have a HUGE application "develoopped" ;) by a big team. Here is a simplified model of the potential disaster that may occur when somebody checks too deep in a data structure. If not possible to disable autovification completely or in scope, how to work around this? Thank you very much :) !!!! use strict; use warnings;use Data::Dumper; my $some_ref = {akey=>{deeper=>1}}; print Dumper($some_ref ); if($some_ref->{deep}{doot} == 1){ print 'too deep '.$/; } if($some_ref->{deep}){ print 'Already in a deep doot'.$/; } print Dumper($some_ref ); This outputs the following: $VAR1 = { 'akey'

This question already has an answer here: What is the best way to implement nested dictionaries? 20 answers Both Google and the online docs are not delivering much insight on my query, so I thought I would ask the community here. In Perl, you can easily setup a hash-of-a-hash-of-a-hash and test the final key like so: my $hash = {}; $hash{"element1"}{"sub1"}{"subsub1"} = "value1"; if (exists($hash{"element1"}{"sub1"}{"subsub1"})) { print "found value\n"; } What's the 'best-practice' equivalent in Python? The closest equivalent is probably something like the following: import collections def

If I understand correctly , calling if (exists $ref->{A}->{B}->{$key}) { ... } will spring into existence $ref->{A} and $ref->{A}->{B} even if they did not exist prior to the if ! This seems highly unwanted. So how should I check if a "deep" hash key exists? It's much better to use something like the autovivification module to turn off that feature, or to use Data::Diver . However, this is one of the simple tasks that I'd expect a programmer to know how to do on his own. Even if you don't use this technique here, you should know it for other problems. This is essentially what Data::Diver is

Is there any way simpler than if hash.key?('a') hash['a']['b'] = 'c' else hash['a'] = {} hash['a']['b'] = 'c' end The easiest way is to construct your Hash with a block argument: hash = Hash.new { |h, k| h[k] = { } } hash['a']['b'] = 1 hash['a']['c'] = 1 hash['b']['c'] = 1 puts hash.inspect # "{"a"=>{"b"=>1, "c"=>1}, "b"=>{"c"=>1}}" This form for new creates a new empty Hash as the default value. You don't want this: hash = Hash.new({ }) as that will use the exact same hash for all default entries. Also, as Phrogz notes, you can make the auto-vivified hashes auto-vivify using default_proc :