ruby methods that either yield or return Enumerator

问题

in recent versions of Ruby, many methods in Enumerable return an Enumerator when they are called without a block:

[1,2,3,4].map 
#=> #<Enumerator: [1, 2, 3, 4]:map> 
[1,2,3,4].map { |x| x*2 }
#=> [2, 4, 6, 8] 

I want do do the same thing in my own methods like so:

class Array
  def double(&block)
    # ???
  end
end

arr = [1,2,3,4]

puts "with block: yielding directly"
arr.double { |x| p x } 

puts "without block: returning Enumerator"
enum = arr.double
enum.each { |x| p x }

回答1:

The core libraries insert a guard return to_enum(:name_of_this_method, arg1, arg2, ..., argn) unless block_given?. In your case:

class Array
  def double
    return to_enum(:double) unless block_given?
    each { |x| yield 2*x }
  end
end

>> [1, 2, 3].double { |x| puts(x) }
2
4
6 
>> ys = [1, 2, 3].double.select { |x| x > 3 } 
#=> [4, 6]

回答2:

use Enumerator#new:

class Array
  def double(&block)
    Enumerator.new do |y| 
      each do |x| 
        y.yield x*2 
      end 
    end.each(&block)
  end
end

回答3:

Another approach might be:

class Array
    def double(&block)
        map {|y| y*2 }.each(&block)
    end
 end

回答4:

easiest way for me

class Array
  def iter
      @lam = lambda {|e| puts e*3}
      each &@lam
  end
end

array = [1,2,3,4,5,6,7]
array.iter

=> 3 6 9 12 15 18 21

来源：https://stackoverflow.com/questions/7183360/ruby-methods-that-either-yield-or-return-enumerator