From: Dominik Bathon <dbatml@...>
Date: 2006-08-17T02:55:55+09:00
Subject: Re: block equality

Hi,

On Wed, 16 Aug 2006 17:21:39 +0200, Tom Jordan <tdjordan@gmail.com> wrote:

> This http://ruby-doc.org/core/classes/Proc.html#M001161
>
> "Return true if prc is the same object as other_proc, or if they are
> both procs with the same body."
>
> led me to believe that Procs would evaluate to equality if they had
> the same body.
>
> So I would have expected that the following:
>
> s = lambda { x = 1 }
> t = lambda { x = 1 }
> s == t   # => true expected, false in reality

If you really want to check if procs have the same or equivalent bodies,  
you can use RubyNode (http://rubynode.rubyforge.org/):

>> p1 = proc { 1 + 1 }
=> #<Proc:0xb7a70b64@(irb):1>
>> p2 = proc { 1 + 1 }
=> #<Proc:0xb7a6e580@(irb):2>
>> require "rubynode"
=> true
>> p1.body_node.transform
=> [:call, {:args=>[:array, [[:lit, {:lit=>1}]]], :mid=>:+, :recv=>[:lit,  
{:lit=>1}]}]
>> p1.body_node.transform == p2.body_node.transform
=> true
>> p1 == p2
=> false

But please be aware that the equality of the body node does not imply that  
the blocks have the same closure:

>> def foo(a) proc { a } end
=> nil
>> p1 = foo(1)
=> #<Proc:0xb7a6217c@(irb):8>
>> p2 = foo(2)
=> #<Proc:0xb7a6217c@(irb):8>
>> p1.body_node.transform == p2.body_node.transform
=> true
>> p1[]
=> 1
>> p2[]
=> 2

And it is also possible that different Ruby code parses to the same node  
tree:

>> p1 = proc { if 1 then 2 else 3 end }
=> #<Proc:0xb7ac83b4@(irb):17>
>> p2 = proc { unless 1 then 3 else 2 end }
=> #<Proc:0xb7ac239c@(irb):18>
>> p3 = proc { 1 ? 2 : 3 }
=> #<Proc:0xb7abec4c@(irb):19>
>> p1.body_node.transform == p2.body_node.transform
=> true
>> p1.body_node.transform == p3.body_node.transform
=> true
>> p2.body_node.transform == p3.body_node.transform
=> true
>> p1.body_node.transform
=> [:if, {:else=>[:lit, {:lit=>3}], :cond=>[:lit, {:lit=>1}],  
:body=>[:lit, {:lit=>2}]}]


So this is probably not very useful for your once method, but interesting  
anyway.

Dominik