From: dblack@...
Date: 2007-09-05T03:36:25+09:00
Subject: Re: assigning to hash keys when there is a default value?

Hi --

On Wed, 5 Sep 2007, Rick DeNatale wrote:

> On 9/3/07, Robert Klemme <shortcutter@googlemail.com> wrote:
>> 2007/9/3, dblack@wobblini.net <dblack@wobblini.net>:
>
>>> x ||= y is, I think, always supposed to be exactly equivalent to
>>> x = x || y, ...
>
>> I can't point my finger on it but I believe x||=y is equivalent to
>> "x=y unless x" instead of "x=x||y". It seems to be more reasonable to
>> skip the assignment altogether if the value is true equivalent
>> already. That would also explain behavior much better. :-)
>
> Robert,
>
> Although I can't find the documentation quickly, although I'm 95%
> certain that it should be in the pickaxe somewhere, I'm pretty sure
> that you are correct.
>
> I've just looked at parse.y and eval.c for ruby1.8.6 and it would appear that:
>
>  h[2] ||= 10
>
> gets compiled to a NODE_OP_ASGN_OR node with h[2] as the lhs and 10 as
> the rhs.  Here's the code from eval.c which evaluates such a node:
>
>      case NODE_OP_ASGN_OR:
> 	if ((node->nd_aid && !is_defined(self, node->nd_head, 0)) ||
> 	    !RTEST(result = rb_eval(self, node->nd_head))) {
> 	    node = node->nd_value;
> 	    goto again;
> 	}
> 	break;
>
> So what happens is that the lhs is only evaluated if the  the lhs
> (node->nd_head) is not defined || it evaluates to an untrue value.
>
> In the case of h[5] the default value for the hash means that it will
> evaluate to 5, and the assignment is not done.
>
> I for one, am glad that it works this way.  The ruby idiom
>
>  x ||= y
>
> is heavily used for lazy initialization/caching.  While most often,
> it's the rhs which is expensive to compute and therefore the thing we
> want to short-circuit, since x= can in general be a method, and might
> just be expensive, then optimizing the case where it boils down to x =
> x as a nop, makes sense.

I think The method always gets called, though:

   class C
     attr_reader :x
     def x=(n)
       puts "C#x="
       true
     end
   end

   c = C.new
   c.x ||= 3    #   C#x=

*Unless*, of course, the object is a Hash which has either (a) a key
corresponding to the indicated value, or (b) a default value with
boolean truth value.

Sigh. I really wish it were otherwise. What an annoying exception to
the rule.

Also, in the famous:

   h = Hash.new(1)
   h[5] ||= 10

case, it definitely isn't doing the "x = x" equivalent, since that
would set the 5 key to 1.

I don't know.... However many times I look at it, I just can't see
this:

   h[5] ||= 10

as *not* meaning that I expect h to end up having a 5 key, one way or
another.


David

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