From: Joel VanderWerf <vjoel@...>
Date: 2009-06-03T07:10:38+09:00
Subject: Re: return in iterator in lambda

Benjamin Ter kuile wrote:
> I want to create a conditional mathematical function as a lambda
> structure. Therefore it may be the case that inside an iterative loop,
> the result of the function is known and should be returned. By
> implementing this it causes trouble. The test code I created looks like:
> 
> 
> func = lambda do
>   for z in [1,2,3]
>     return 2
>   end
>  "something"
> end
> puts func.call
> puts "after call"
> 
> This results in a LocalJumpError (Ruby 1.8.6 and 1.8.7). In ruby 1.9 it
> works as expected. Is there a proper way of creating a function like
> this in ruby < 1.9?

Enumerable#find may be more appropriate than #each. This avoids the 
issue of how to break out of nested blocks.

func = lambda do |a|
   a.find do |z|
     z if z >= 1.5
   end or "not found"
end

puts func.call([1,2,3])
puts func.call([0,0.2,0.4])

-- 
       vjoel : Joel VanderWerf : path berkeley edu : 510 665 3407