From: Robert Klemme <shortcutter@...>
Date: 2007-07-31T18:45:30+09:00
Subject: Re: how to make a[2][3]=4 work?

2007/7/23, Fsormok Fsormok <fsormok@inbox.ru>:
> > a = Hash.new{|h,k| h[k]=Hash.new &h.default_proc}
>
> > irb(main):026:0> require 'ostruct'
> > irb(main):027:0> int = Hash.new{|h,k| h[k]=OpenStruct.new}
>
> >a = {}
> >    ==>{}
> >a[1] = 1
> >    ==>1
> >a[ [2,1] ] = 2
> >    ==>2
> >a[ [2,2,3] ] = 4
> >    ==>4
> >p a
> >{[2, 2, 3]=>4, [2, 1]=>2, 1=>1}
> >    ==>nil
>
> Tanks for the replies.
>
> I'm just a little bit confused because in ruby everything is lean and
> clean but when it comes to hashes of hashes it gets kind of complicated.
> For me it looks like ruby has another "concept" of solving such things.

Its this what you are looking for?

$ ruby <<XXX
> miss = lambda {|h,k| h[k] = Hash.new(&miss)}
> hash = Hash.new(&miss)
> hash[1][2]=3
> hash[4]=5
> p hash
> XXX
{1=>{2=>3}, 4=>5}

Kind regards

robert