From: Julian Leviston <julian@...>
Date: 2013-03-30T01:39:55+09:00
Subject: Re: Hash with default


On 30/03/2013, at 3:02 AM, Adam Prescott <adam@aprescott.com> wrote:

> On 29 March 2013 15:53, Julian Leviston <julian@coretech.net.au> wrote:
>> h = {"two"=>2,"three"=>3}; def h.[](key); self.has_key?(key) ? super(key) : key; end
> 
> Why not use the block to Hash.new instead of overriding #[] ?
> 
>>> h = Hash.new { |h,k| k }
> => {}
>>> h["one"]
> => "one"
>>> h
> => {}
> 
> Note this is different than using h[k] = k in the block.
> 
> As for the original question, I think you can rely on default_proc and
> get around the fact it doesn't return the receiver pretty easily.
> 
>>> h = { "two" => 2, "three" => 3 }.tap { |o| o.default_proc = lambda { |h, k| k } }
> => {"two"=>2, "three"=>3}
>>> h
> => {"two"=>2, "three"=>3}
>>> h["two"]
> => 2
>>> h["five"]
> => "five"
>>> h
> => {"two"=>2, "three"=>3}
> 
> Having said that, I don't see any benefit in trying to get it down to
> a single line and sacrificing readability. You can define h to be what
> you want, then on a separate line just call #default_proc=.
> 

Sorry. I apologise. My information was outdated, and I stand corrected!