From: Robert Klemme <shortcutter@...>
Date: 2009-03-01T07:33:58+09:00
Subject: Re: Newbie question > Hash workings

On 28.02.2009 22:15, AD60 wrote:
> I am puzzled by the workings a the following program. I placed the
> output after the corresponding lines. To me it lookes like a sort of
> bug in the workings of de p method. Can anybody set me straight?

p works correctly but...

> h = Hash.new([])
> h[0] = [1,2,3]
> p h                      #> {0=>[1, 2, 3]}
> h[0] += [4]
> p h                      #> {0=>[1, 2, 3, 4]}
> h[1] += [10,11]
> p h                      #> {0=>[1, 2, 3, 4], 1=>[10, 11]}
> h[1] << 12
> p h                     #> {0=>[1, 2, 3, 4], 1=>[10, 11, 12]}
> h[2] << 20
> p h                     #> {0=>[1, 2, 3, 4], 1=>[10, 11,
> 12]                          ?
> p h[2]                 #>
> [20]                                                                   ???
> h[2] << 21
> p h                     #> {0=>[1, 2, 3, 4], 1=>[10, 11,
> 12]}                          ?
> h[2] += [22]
> p h                     #> {0=>[1, 2, 3, 4], 1=>[10, 11, 12], 2=>[20,
> 21, 22]}  ????????

... you are modifying the default value of the Hash in those cases where 
you place the question marks.  The default value is the one returned in 
case a key does not exist.  But there is no automatic insertion, that's 
why you do not see key 2 after doing h[2] << 20.

You probably rather want this famous idiom:

h = Hash.new {|ha,ke| ha[ke] = []}

Now run your tests again. :-)

Kind regards

	robert