From: Eric Mahurin <eric_mahurin@...>
Date: 2005-10-02T08:37:37+09:00
Subject: Re: ideas for an RCR: variable locality

--- Bob Hutchison <hutch@recursive.ca> wrote:
> On Oct 1, 2005, at 4:55 PM, Eric Mahurin wrote:
> > Here's one - a poor man's macro facility.  Let's say a
> macro is
> > just a lambda that returns a string.  You just eval it when
> you
> > need to execute the code for that macro.
> >
> > plus  = lambda { |a,b| "(#{a}+#{b})" }
> > # will have to re-evaluate a or b if they are an expression
> > min1  = lambda { |a,b| "(#{a}<#{b} ? #{a} : #{b})" }
> > # use local variables to prevent re-evaluation
> > min2  = lambda { |a,b| "(a=#{a};b=#{b};a<b ? a : b)" }
> >
> > # y+z may get evaluated twice
> > min1["x",plus["y","z"]]
> > # => "(x<(y+z) ? x : (y+z))"
> >
> > # y+z may evaluated once
> > min2["x",plus["y","z"]]
> > # => "(a=x;b=(y+z);a<b ? a : b)"
> >
> > # need to localize a/b for the inner min2
> > min2["x",min2["y","z"]]
> > # => "(a=x;b=(a=y;b=z;a<b ? a : b);a<b ? a : b)"
> >
> >
> > See the problem on this last example?  We really need to
> > localize a and b.  There isn't a good facility to do this.
> >
> >
> 
> You need gensym as lisp has (and I had in a previous
> example), in  
> which case you get this back on the last example, and there
> is no  
> problem:
> 
> (a_1=x;b_2=(a_3=y;b_4=z;a_3<b_4 ? a_3 : b_4);a_1<b_2 ? a_1 :
> b_2)

Yes, something like this would work (using your Gensym module):

min2  = lambda { |a,b|
  v1 = Gensym.gensym
  v2 = Gensym.gensym
  "(#{v1}=#{a};#{v2}=#{b};#{v1}<#{v2} ? #{v1} : #{v2})"
}

Here are the disadvantages of this compared to having ruby do
local variables:

- not as natural/uglier
- clutters the namespace/variable table
- prevents GC from freeing objects that these variables
reference




		
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