Re: Hash with default

On 30/03/2013, at 3:02 AM, Adam Prescott <[e-mail]> wrote:

> On 29 March 2013 15:53, Julian Leviston <[e-mail]> =
wrote:
>> h =3D {"two"=3D>2,"three"=3D>3}; def h.[](key); self.has_key?(key) ? =
super(key) : key; end
>=20
> Why not use the block to Hash.new instead of overriding #[] ?
>=20
>>> h =3D Hash.new { |h,k| k }
> =3D> {}
>>> h["one"]
> =3D> "one"
>>> h
> =3D> {}
>=20
> Note this is different than using h[k] =3D k in the block.
>=20
> As for the original question, I think you can rely on default_proc and
> get around the fact it doesn't return the receiver pretty easily.
>=20
>>> h =3D { "two" =3D> 2, "three" =3D> 3 }.tap { |o| o.default_proc =3D =
lambda { |h, k| k } }
> =3D> {"two"=3D>2, "three"=3D>3}
>>> h
> =3D> {"two"=3D>2, "three"=3D>3}
>>> h["two"]
> =3D> 2
>>> h["five"]
> =3D> "five"
>>> h
> =3D> {"two"=3D>2, "three"=3D>3}
>=20
> Having said that, I don't see any benefit in trying to get it down to
> a single line and sacrificing readability. You can define h to be what
> you want, then on a separate line just call #default_proc=3D.
>=20

Sorry. I apologise. My information was outdated, and I stand corrected!