From: jmichel@... (Jean Michel)
Date: 2001-01-04T05:10:02+09:00
Subject: [ruby-talk:8589] Re: bug??

In article <Pine.GSO.4.21.0101031839410.4268-100000@neelix>,
Hugh Sasse Staff Elec Eng  <hgs@dmu.ac.uk> wrote:
>On Thu, 4 Jan 2001, Jean Michel wrote:
>

Thank you  very much  for the  explanation. But there  are two  things I
don't completely understand yet (or do not like - they do not seem to me
to follow  the principle of  least surprise,  and they make  perl idioms
fail) about the behaviour:

>> irb(main):015:0> a=Hash.new("")
>> {}
>
>Creates the new hash, and a single object <""> to return when 
>an nonexistent element is accessed.
>
>> irb(main):016:0> a[1]<<"a"
>> "a"
>
>Access a nonexistent element -> ""
>append "a" to it :  "" << "a" -> "a"
>
>> irb(main):017:0> a[1]
>> "a"
>
>Access a non existenet element.  Get the one object back, as
>before, but you just modified it, so it is "a", not "" any more.
>
>> irb(main):018:0> a
>> {}
>
>You still haven't defined any keys for the hash yet, so there
>is nothing in it.

The first thing I do not like is that a[1]<<"a" does not create the key.
I would  expect that any method  which modifes the receiver  creates the
key. I guess here to define the key the only solution is a[1]<<="a" ?

The second  thing I  do not  like is that  when creating  a non-existent
element,  ruby  does  not  do  a  "dup" to  give  a  new  copy  of  this
element. This  more or less  destroys the  usefulness of the  perl idiom
'automagically create elements'. Indeed if you do

irb(main):019:0> a=Hash.new("")
{}
irb(main):020:0> a[1]<<="a"
"a"
irb(main):021:0> a
{1=>"a"}
irb(main):022:0> a[2]<<="b"
"ab"
irb(main):023:0> a
{1=>"ab", 2=>"ab"}

while in perl you would get

{1=>"a", 2=>"b"}

Is there any way to get the perl behaviour?

Best regards,
   Jean MICHEL