module I18n::Backend::Pluralization

def pluralize(locale, entry, count)
  return entry unless entry.is_a?(Hash) && count
  pluralizer = pluralizer(locale)
  if pluralizer.respond_to?(:call)
    # Deprecation: The use of the `zero` key in this way is incorrect.
    # Users that want a different string for the case of `count == 0` should use the explicit "0" key instead.
    # We keep this incorrect behaviour for now for backwards compatibility until we can remove it.
    # Ref: https://github.com/ruby-i18n/i18n/issues/629
    return entry[:zero] if count == 0 && entry.has_key?(:zero)
    # "0" and "1" are special cases
    # https://unicode-org.github.io/cldr/ldml/tr35-numbers.html#Explicit_0_1_rules
    if count == 0 || count == 1
      value = entry[symbolic_count(count)]
      return value if value
    end
    # Lateral Inheritance of "count" attribute (http://www.unicode.org/reports/tr35/#Lateral_Inheritance):
    # > If there is no value for a path, and that path has a [@count="x"] attribute and value, then:
    # > 1. If "x" is numeric, the path falls back to the path with [@count=«the plural rules category for x for that locale»], within that the same locale.
    # > 2. If "x" is anything but "other", it falls back to a path [@count="other"], within that the same locale.
    # > 3. If "x" is "other", it falls back to the path that is completely missing the count item, within that the same locale.
    # Note: We don't yet implement #3 above, since we haven't decided how lateral inheritance attributes should be represented.
    plural_rule_category = pluralizer.call(count)
    value = if entry.has_key?(plural_rule_category) || entry.has_key?(:other)
      entry[plural_rule_category] || entry[:other]
    else
      raise InvalidPluralizationData.new(entry, count, plural_rule_category)
    end
  else
    super
  end
end

def pluralizer(locale)
  pluralizers[locale] ||= I18n.t(:'i18n.plural.rule', :locale => locale, :resolve => false)
end

def pluralizers
  @pluralizers ||= {}
end

def symbolic_count(count)
  count = 0 if count == 0
  count = 1 if count == 1
  count.to_s.to_sym
end