如何实现咖喱红宝石（部分功能）

Question

我需要实现在红宝石咖喱功能（1.8.6或1.8.7不1.9）的一些示例。

Answer 1

因此，这里是如何做到与块讨好，而不是方法：

def curry(&block) 
  arity = (block.arity >= 0) ? block.arity : -(block.arity + 1)
  # return an immediate value if the block has one
  return block[] if arity == 0

  # otherwise, curry it argument by argument
  args = []
  innermost = lambda do |last,*extra|
    args[arity-1] = last
    block[*(args+extra)]
  end
  (0...(arity-1)).to_a.reverse.inject(innermost) do |inner,i|
    lambda do |arg_i,*extra|
      args[i] = arg_i
      # pass extra arguments on to inner calls
      if extra.empty?
        inner
      else
        inner[*extra]
      end
    end
  end
end

和它的作品相当不错的做法。参数可以令行禁止或没有， 额外的参数被收集照例：

irb> (curry { |x,y| x + y })[1,2]
#=> 3
irb> (curry { |x,y| x + y })[1][2]
#=> 3
irb> (curry { |x,*ys| ys << x })[1]
#=> [1]
irb> (curry { |x,*ys| ys << x })[1,2,3]
#=> [2, 3, 1]
irb> (curry { |x,y,*zs| zs << (x+y) })[1,2]
#=> [3]
irb> (curry { |x,y,*zs| zs << (x+y) })[1,2,4]
#=> [4, 3]
irb> (curry { |x,y,*zs| zs << (x+y) })[1][2]
#=> [3]
irb> (curry { |x,y,*zs| zs << (x+y) })[1][2,4]
#=> [4, 3]
irb> (curry { |a,b,c,d,e| a+b+c+d+e })[1,2,3,4,5]
#=> 15
irb> (curry { |a,b,c,d,e| a+b+c+d+e })[1][2][3][4][5]
#=> 15
irb> (curry { |a,b,c,d,e| a+b+c+d+e })[1,2][3][4][5]
#=> 15
irb> (curry { |a,b,c,d,e| a+b+c+d+e })[1][2,3,4][5]
#=> 15

我作出的设计决策已经没有精氨酸块上讨好返回一个立即值：

irb> curry { 3 }
#=> 3
irb> curry { |*xs| xs }
#=> []

这是必要的，以避免不得不结束与[]每次讨好（和相当的Haskell样）。