From: Rick DeNatale <rick.denatale@...>
Date: 2009-05-06T21:03:43+09:00
Subject: Re: '=||'

On Wed, May 6, 2009 at 6:54 AM, Eleanor McHugh
<eleanor@games-with-brains.com> wrote:
> On 6 May 2009, at 02:31, Rick DeNatale wrote:
>>
>> On Tue, May 5, 2009 at 8:45 PM, Eleanor McHugh
>> <eleanor@games-with-brains.com> wrote:
>>>
>>> On 6 May 2009, at 00:09, 7stud -- wrote:
>>>>
>>>> The statement:
>>>>
>>>> x ||= val
>>>>
>>>> is actually equivalent to:
>>>>
>>>> x = val unless x
>>>
>>>
>>> It seems you've misunderstood what happens under the hood when using
>>> augmented assignment with tables as '||=' then becomes syntactic sugar
>>> for
>>> 'x[] = x[] || some_other_value' and the assignment is performed via '[]='
>>> rather than '='. '[]=' will not create a key if it believes it already
>>> exists and this is the cause of the behaviour you're seeing.
>>
>> No, if x is truthy then
>>
>>  x ||= expression
>>
>> will NOT do any assignment.
>>
>> The real equivalent to x ||= y
>>
>> is
>>
>> x || x = y
>>
>> The assignment is short-circuited.
>>
>> For the proof see:
>>
>> http://talklikeaduck.denhaven2.com/2008/04/26/x-y-redux
>
> Yes, for assignment that's the case. But 'x[n] ||= y' isn't an instance of
> assignment in that case as can easily be demonstrated:
>
> class Test
>  def method_missing symbol, *args
>    puts "calling method #{symbol}"
>  end
> end
>
> t = Test.new
> t[:a] ||= 17
>
> output: calling method []
>                calling method []=
>
> Notice how even though method_missing returns a value and is thus 'true' the
> sequence still attempts all parts of the expression, and
>
> t[:a] = t[:a] || 17
>
> output: calling method []
>                calling method []=
>                => 17
>
> confirms that no short-circuited evaluation occurs.
>

But this is because the call to the missing [] method goes to the
method missing method which returns nil.

Remember that the assertion is that

t[:a] ||= 17

is equivalent to

(t[:a]) || (t[:a] = 17)

which in turn is equivalent to:

((t.[](:a)) || (t.[]=(:a, 17))

The output of your example shows both the :[] and :[]= methods are being sent.

Try this variant:

class Test
 def method_missing symbol, *args
   puts "calling method #{symbol}"
 end

 def [](a)
   puts "in [] method"
   a
 end
end

t = Test.new
t[:a] ||= 17

This produces the output;
in [] method

The short circuiting only happens if the lhs expression returns a
non-truthy value.

Now it's true (I think) that "a op= b" is the same as "a = a op b" in
C, Ruby ain't C.

Sometimes the stuff "under the hood" is a little more complicated than
it first appears. <G>

-- 
Rick DeNatale

Blog: http://talklikeaduck.denhaven2.com/
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