As mentioned in this answer, Array.new(size, object) creates an array with size references to the same object.
hash = Hash.new
a = Array.new(2, hash)
a[0]['cat'] = 'feline'
a # => [{"cat"=>"feline"},{"cat"=>"feline"}]
a[1]['cat'] = 'Felix'
a # => [{"cat"=>"Felix"},{"cat"=>"Felix"}]
Why does Ruby do this, rather than doing a dup or clone of object?
Because that's what the documentation says it does. Note that Hash.new is only being called once, so of course there's only one Hash
If you want to create a new object for every element in the array, pass a block to the Array.new method, and that block will be called for each new element:
>> a = Array.new(2) { Hash.new }
=> [{}, {}]
>> a[0]['cat'] = 'feline'
=> "feline"
>> a
=> [{"cat"=>"feline"}, {}]
>> a[1]['cat'] = 'Felix'
=> "Felix"
>> a
=> [{"cat"=>"feline"}, {"cat"=>"Felix"}]
Array.new is indeed the parameter used to pre-populate the array. Sarcastic, because Gareth said as much in his answer, and I am simply restating what he said.
For certain classes that can't be modified in-place (like Fixnums) the Array.new(X, object) form works as expected and is probably more efficient (it just calls memfill instead of rb_ary_store and yielding to the block):
For more complicated objects you always have the block form (e.g. Array.new(5) { Hash.new }).
*Edit:* Modified according to the comments. Sorry for the stupid example, I was tired when I wrote that.
I came up with this answer, very short and simple solution
c_hash = Hash.new
["a","b","c","d","e","f"].each do |o|
tmp = Hash.new
[1,2,3].map {|r| tmp[r] = Array.new}
c_hash[o] = tmp
end
c_hash['a'][1] << 10
p c_hash
