rubyarrayshashenumerable

Why does Enumerable#find/#detect return an Array even when called on an Hash?

The documentation for Enumerable#find/#detect says:

find(ifnone = nil) { |obj| block } → obj or nil
find(ifnone = nil) → an_enumerator

Passes each entry in enum to block. Returns the first for which block is not false. If no object matches, calls ifnone and returns its result when it is specified, or returns nil otherwise.

However, when it is called on the Hash, the result has changed the type to Array instead of the original Hash.

Is it some implementation fault or some historical conventions regarding this datatype?

{a: 'a', b:'b'}.find {|k, v| v == 'b'}
# => [:b, 'b']
Solution

  • The Hash#detect is inherited from Enumerable#detect method.

    Enumerable module generates multiple methods(such as sort, min, max including detect etc.) based on the each method of the class which includes Enumerable.

    It doesn't care about how each is implemented as long as it

    "...yields successive members of the collection. " from ruby-doc

    So for the Hash#detect method, it relies on Hash#each's behavior, which is:

    Calls block once for each key in hsh, passing the key-value pair as parameters. If no block is given, an enumerator is returned instead.

    h = { "a" => 100, "b" => 200 }
h.each {|key, value| puts "#{key} is #{value}" }

    Because Hash#each yields the hash as two pair array, all methods inherited from the Enumerable module works based on that.

    That's why Hash#detect produces a two elements array instead of the an hash object itself.