是否有汽车,cdr和缺点的红宝石？

Question

是否有红宝石等效的lisp汽车,cdr和cons功能？对于那些不熟悉lisp的人来说,这就是我想要的ruby:

[1,2,3].car   => 1
[1,2,3].cdr   => [2,3]
[2,3].cons(1) => [1,2,3]

(在口齿不清):

(car '(1 2 3))  => 1
(cdr '(1 2 3))  => (2 3)
(cons 1 '(2 3)) => (1 2 3)

Answer 1

Ruby数组并没有实现为单链接列表,因此拥有car和cdr并不是很有用.

如果你真的想要,你可以做到

[1,2,3][0]      => 1
[1,2,3].first   => 1
[1,2,3][1..-1]  => [2,3]
[1] + [2,3]     => [1,2,3]

Answer 2

这就是你如何在ruby中实现类似lisp的单链表:

class Object
  def list?
    false
  end
end

class LispNilClass
  include Enumerable
  def each
  end

  def inspect
    "lnil"
  end

  def cons(car)
    Cell.new(car, self)
  end

  def list?
    true
  end
end

LispNil = LispNilClass.new

class LispNilClass
  private :initialize
end

class Cell
  include Enumerable

  attr_accessor :car, :cdr

  def initialize(car, cdr)
    @car = car
    @cdr = cdr
  end

  def self.list(*elements)
    if elements.empty?
      LispNil
    else
      first, *rest = elements
      Cell.new(first, list(*rest))
    end
  end

  def cons(new_car)
    Cell.new(new_car, self)
  end

  def list?
    cdr.list?
  end

  # Do not use this (or any Enumerable methods) on Cells that aren't lists
  def each
    yield car
    cdr.each {|e| yield e}
  end

  def inspect
    if list?
      "(#{ to_a.join(", ") })"
    else
      "(#{car} . #{cdr})"
    end
  end
end

list = Cell.list(1, 2, 3) #=> (1, 2, 3)
list.list? #=> true
list.car #=> 1
list.cdr #=> (2, 3)
list.cdr.cdr.cdr #=> lnil
list.cons(4) #=> (4, 1, 2, 3)

notlist = Cell.new(1,2) #=> (1 . 2)
notlist.list? #=> false
notlist.car #=> 1
notlist.cdr #=> 2
notlist.cons(3) #=> (3 . (1 . 2))

Answer 3

半严肃的说,如果你想在Ruby中使用CONS,CAR和CDR,你可能会比这更糟糕

def cons(x,y)
   return lambda {|m| m.call(x,y)}
end

def car(z)
  z.call(lambda {|p,q| p})
end

def cdr(z)
  z.call(lambda {|p,q| q})
end

然后你可以定义你的列表程序,

def interval(low, high)
  if (low > high)
    return nil
  else
    return cons(low, interval(low + 1, high))
  end
end

def map(f, l)
  if (l == nil)
    return nil
  else
    cons(f.call(car(l)), map(f, cdr(l)))
  end
end

def filter(p, l)
  if (l == nil)
    return nil
  elsif (p.call(car(l)))
    return cons(car(l), filter(p, cdr(l)))
  else
    return filter(p, cdr(l))
  end
end

def reduce(f, f0, l)
  if (l == nil)
    return f0
  else
    return f.call(car(l), reduce(f, f0, cdr(l)))
  end
end

然后你可能得到1到10范围内的奇数平方的总和:

reduce(lambda {|x, y| x + y},
       0,
       filter(lambda {|x| x % 2 == 1},
              map(lambda {|x| x * x},
                  interval(1, 10))))
=> 165

Answer 4

>> [1,2,3].drop 1
=> [2, 3]
>> [1,2,3].first
=> 1

当然,如你所知,这些与Lisp并不太接近.真正的红宝石等价物就像是[1, [2, [3, nil]]].你总是可以写一个List类......或者在某个地方找一个.

实用Ruby项目的第8章在Ruby中称为实现Lisp.