Hash.new(confusion)

[Trans]

irb(main):006:0> def r
irb(main):007:1> @r ||= Hash.new({})
irb(main):008:1> end
=> nil
irb(main):009:0> r[:a]
=> {}
irb(main):010:0> r[:b][:c] = 10
=> 10
irb(main):011:0> r
=> {}
irb(main):012:0> r[:b]
=> {:c=>10}
irb(main):013:0>

Please explain how r is empty but r[:b] exists. Thank You.

T.

 

[Joel VanderWerf]

irb(main):006:0> def r
irb(main):007:1> @r ||= Hash.new({})
irb(main):008:1> end
=> nil
irb(main):009:0> r[:a]
=> {}
irb(main):010:0> r[:b][:c] = 10
=> 10
irb(main):011:0> r
=> {}
irb(main):012:0> r[:b]
=> {:c=>10}
irb(main):013:0>

Please explain how r is empty but r[:b] exists. Thank You.

T.

Try

p r.default

and you'll see what's going on.

 

[Gennady Bystritsky]

If you specify a parameter for Hash.new the SAME object will be used
for any default value (see documentation). If you modify it, it will
be modified for all default values afterwards. Try to see what the
value will be returned for r[:a] if you do it again. I bet it will be
{:c => 10}. As well as for r[:c], r[10] and so on.

Gennady.

 

[Mitch Tishmack]

r is independent of r[:b]. r[:b] is a hash with a value containing
another hash/key value.

ie {:b => {:c => 10}} NOT {:b => 10, :c = 10}. If I understand your
confusion correctly.

irb(main):001:0> def r
irb(main):002:1> @r ||= Hash.new({})
irb(main):003:1> end
=> nil
irb(main):004:0> r[:a]
=> {}
irb(main):005:0> r.empty?
=> true
irb(main):006:0> r[:a] = 1
=> 1
irb(main):007:0> r.empty?
=> false
irb(main):008:0> r[:b][:c] = 10
=> 10
irb(main):009:0> r
=> {:a=>1}
irb(main):010:0> r[:b]
=> {:c=>10}
irb(main):011:0> (r[:b])[:c]
=> 10
irb(main):012:0> r[:a]
=> 1
irb(main):013:0> r.size
=> 1
irb(main):014:0> r[:b].size
=> 1
irb(main):015:0> r[:b][:d] = 300000
=> 300000
irb(main):016:0> r[:b].size
=> 2
irb(main):018:0> r
=> {:a=>1}
irb(main):019:0> r[:b]
=> {:d=>300000, :c=>10}

Or did I completely miss your point?

Mitch

 

[Dave Burt]

...
Please explain how r is empty but r[:b] exists. Thank You.

# To summarize:
h = Hash.new({})
h[1][2] = 3
h #=> {}
h.default #=> {2=>3}

So you've added to the default object. It might be helpful to note another
wrong way to do this:
h = Hash.new { {} }
h[1][2] = 3
h #=> {}
h.default #=> {}

Different, because default now returns a new hash every time.

But, to answer the implied question, "what is the code that will do what I
intend?" there are a few ways to do this:

# Just one level of magic new hashes
h = Hash.new {|hash, key| hash[key] = {} }
h[1][2] = 3
h #=> {1=>{2=>3}}
h[1][2][3] = 4 # Error (calling []= on nil)

# Auto-vivifying hash: using proc
auto_vivivy_hash = proc {|hash, key| hash[key] = Hash.new
&auto_vivivy_hash }
h = Hash.new &auto_vivify_hash
h[1][2][3] = 4 #=> {1=>{2=>{3=>4}}}

# Auto-vivifying hash: using Hash subclass
class X < Hash
def default(key = nil)
self[key] = self.class.new
end
end
h = X.new #=> {}
h[1][2][3] = 4
h #=> {1=>{2=>{3=>4}}}

Cheers,
Dave

 

[David A. Black]

Hi --

irb(main):006:0> def r
irb(main):007:1> @r ||= Hash.new({})
irb(main):008:1> end
=> nil
irb(main):009:0> r[:a]
=> {}
irb(main):010:0> r[:b][:c] = 10
=> 10
irb(main):011:0> r
=> {}
irb(main):012:0> r[:b]
=> {:c=>10}
irb(main):013:0>

Please explain how r is empty but r[:b] exists. Thank You.

r[:b] doesn't exist, so instead, you get the hash's default value
(which is what is returned for non-existent keys).


David

 

[Trans]

Ahhhh, Now I _finally_understand. It's quite subtle.

Thank You All!

T.