From: "Mauricio Fernández" <batsman.geo@...>
Date: 2003-06-13T21:23:17+09:00
Subject: Re: Reason for implicit block syntax ?

On Fri, Jun 13, 2003 at 09:01:59PM +0900, Chris Pine wrote:
> ----- Original Message ----- 
> From: "Mauricio Fern�ndez" <batsman.geo@yahoo.com>
> 
> 
> My bet is that he means
>   my_func bar, baz { ... block ... }
> as opposed to
>   block = proc { ... block .., }
> ----------------------------
> 
> 
> But since `proc' is just a method which takes a block... aren't these
> practically identical syntax?  As I see it, there's really only one way to
> make a block.

You chose the wrong line :-)
The point was that the "normal way" (with lambdas but not blocks � la
Ruby) would be
  my_func(bar, baz, block) 
ie. passing the lambda explicitly and having to use block.call inside
my_func, instead of having yield call the "implicit block".

As for the creation of the block, that's by far the most common (and
sensible) way, but what about (silly example)

irb(main):003:0> class A
irb(main):004:1>   def foo(bar)
irb(main):005:2>     puts "A#foo #{bar}"
irb(main):006:2>   end
irb(main):007:1> end
=> nil
irb(main):008:0> block = A.instance_method(:foo).bind(A.new)
=> #<Method: A#foo>
irb(main):009:0> 2.times &block
(irb):9: warning: `&' interpreted as argument prefix
A#foo 0
A#foo 1
=> 2
irb(main):010:0> block = A.new.method(:foo).to_proc
=> #<Proc:0x40189040@(irb):10>
irb(main):011:0> 2.times &block
(irb):11: warning: `&' interpreted as argument prefix
A#foo 0
A#foo 1
=> 2


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