Nobu [mailto:nobu.nokada / softhome.net]:
> 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)

Where will the ** notation fit in? Will it be:

  foo(** :foo => 14, :bar => 92)
  foo(**{ :foo => 14, :bar => 92 })

And this will effectively make one of those two:

> foo(foo: 14, bar: 92)

?

-austin
--
austin ziegler * austin.ziegler / evault.com