From: Robert Klemme <shortcutter@...>
Date: 2011-04-04T16:36:33+09:00
Subject: Re: pipe question

On Mon, Apr 4, 2011 at 5:08 AM, John Feminella <johnf@bitsbuilder.com> wrote:
> An expression like "foo || bar" works like this: First, evaluate the
> boolean value of `foo`. If it is boolean-true, then the value of this
> expression is `foo`. If not, then the value of this expression is
> `bar`.
>
> This expression says, assign the current value of @first_name back
> into @first_name. Alternatively, if this is the first time you're
> trying to assign it, it'll be nil, so use an empty string instead.
>
> It's equivalent to writing:
>
>    @first_name ||= ''

No.  That is equivalent to:

@first_name || @first_name = ""

which is not the same as

@first_name = @first_name || ""

See also

irb(main):004:0> x=Object.new
=> #<Object:0x10924b1c>
irb(main):005:0> def x.[](k) puts "get";@x;end
=> nil
irb(main):006:0> def x.[]=(k,v) puts "put";@x=v;end
=> nil
irb(main):007:0> x[1] ||= 2
get
put
=> 2
irb(main):008:0> x[1] ||= 2
get
=> 2

This has been discussed at lengths here in the past.  You should be
able to find plenty of evidence. :-)

Kind regards

robert

-- 
remember.guy do |as, often| as.you_can - without end
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