Hi,

At Thu, 3 Jun 2004 02:33:44 +0900,
Paul Brannan wrote in [ruby-talk:102177]:
> But how would (a) work?  It seems like it would make the following
> ambiguous:
> 
> def foo(foo, bar=2)
> end
> 
> foo(:foo => 14, :bar => 92)
> 
> (is this calling foo with a hash and leaving bar with default argument,
> or is it calling foo with foo equal to 14 and bar equal to 92?)

The former.  The latter syntax will be

  foo(foo: 14, bar: 92)

-- 
Nobu Nakada