From: "David A. Black" <dblack@...>
Date: 2009-07-05T10:10:55+09:00
Subject: Re: to_proc and Proc/block conversion with &

Hi --

On Sun, 5 Jul 2009, Russ McBride wrote:

>
> map requires a block, right?  So this works:
>
> words = %w(ardvark   anteater   llama)
>
> arr = words.map { |x| x + "!" }
> arr     #--> [ardvark!, anteater!, llama!]
>
> But this doesn't:
>
> arr2 = words.map lambda{ |x|  x + "!"}       #--> ArgumentError: wrong 
> number of arguments (1 for 0)
>
>
> We can trigger a call to some object's to_proc method with the ampersand. 
> For example, this
> will trigger the symbol's to_proc method, assuming there is one:
>
> upcase_words = words.map(&:upcase)
>
> and if we have already have this:
>
> class Symbol
>   def to_proc
>     lambda{ |x| x.send(self) }
>   end
> end
>
> then we get:
> #--> [ARDVARK, ANTEATER, LLAMA]
>
> But, the to_proc method here is returning a Proc, not a block!  So why does 
> this work?

There's no such thing as returning a block. A block is a syntactic
construct. You can, however, have a Proc object in block position,
playing the block role, courtesy of &. (We don't have a really perfect
term for that, I think.)

> As far as I can tell, the ampersand must be doing 2 things:
> 1- triggering the call to to_proc
> 2- converting the returned Proc object into a block for map to handle.
>
> Is this right?

Yes; the & means: take what's to the right of it (which can be any
expression), evaluate it, call to_proc on the resulting object, and
use the return value of to_proc to play the code block role.

In the case of a lambda, to_proc just returns the receiver. So you can
do:

   array.map &lambda {|x| x * 10 }

Without the &, you're just putting an argument in method-argument
position. Using a Proc object as a method argument does not trigger
any special or block-related behavior. That comes entirely from the
ampersand.


David

-- 
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