From: "David A. Black" <dblack@...>
Date: 2009-08-27T08:23:44+09:00
Subject: Re: ||= with 1.8 and 1.9 ?

Hi --

On Thu, 27 Aug 2009, Yukihiro Matsumoto wrote:

> Hi,
>
> In message "Re: ||= with 1.8 and 1.9 ?"
>    on Thu, 27 Aug 2009 06:23:01 +0900, "David A. Black" <dblack@rubypal.com> writes:
>
> |> According the rubyspec project:
> |>
> |>  language/variables_spec.rb:
> |>    describe "Operator assignment 'var op= expr'" do
> |>      it "is equivalent to 'var = var op expr'" do
> |
> |It isn't, though, at least in the ||= case (see Joel W.'s post).
> |Matz's characterization of it to me at RubyConf (or somewhere) was:
> |
> |   x ||= y   same as    x || x = y
>
> Basically, "var op= expr" works like "var = var op expr" but left hand
> side expression (var) is evaluated only once.  For ||=, it is same as
>
>  x = (x || y)
>
> but
>
>  x || (x = y)
>
> is semantically same (well, almost).

The hash case, though, definitely points to x || (x = y):

>> h = Hash.new(5)
=> {}
>> h[:x] = (h[:x] || 1)
=> 5
>> h
=> {:x=>5}
>> h[:z] || (h[:z] = 1)
=> 5
>> h
=> {:x=>5}
>> h[:y] ||= 1
=> 5
>> h
=> {:x=>5}


David

-- 
David A. Black / Ruby Power and Light, LLC / http://www.rubypal.com
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