From: Jacob Fugal <lukfugl@...>
Date: 2006-01-06T02:13:13+09:00
Subject: Re: Default block parameter?

On 1/4/06, dblack@wobblini.net <dblack@wobblini.net> wrote:
> On Thu, 5 Jan 2006, Mark J.Reed wrote:
>
> > Okay, this is probably a dumb question, but how do I declare an
> > optional block parameter with a default value?
> >
> > I tried several variations on this basic theme:
> >
> >        def meth(&block = lambda { |i| ... })
> >        ...
> >        end
>
> The &block thing is a special dispensation from Ruby, letting you grab
> the block but not serving as a normal argument.  The way I've always
> seen this done is:
>
>    def meth(&block)
>      block ||= lambda { ... }
>      ...
>    end

But keep in mind that assigning to block inside the method doesn't
affect the behavior of yield:

irb> def test(&block)
irb>   block ||= lambda{ puts "default" }
irb>   yield
irb> end
=> nil
irb> test
LocalJumpError: no block given

So if you need a default block and currently use yield, you'll either
need to branch on block_given? (as suggested by James), or just use
block.call instead of yield. The latter is probably preferrable, but
may have subtle differences in parameter assignment if it matters.

Jacob Fugal