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def winner(p1, p2)
  wins = {rock: :scissors, scissors: :paper, paper: :rock}
  {true => p1, false => p2}[wins[p1] == p2]
end

来自这个问题:HW impossibility?:“在 ruby​​ 中创建一个石头剪刀布程序而不使用条件”

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2 回答 2

6

我承认,对于新手程序员来说,这不是最易读的代码。我重写了它,提取了一些变量并添加了注释。希望你现在能更好地理解它。

def winner(p1, p2)
  # rules of dominance, who wins over who
  wins = {
    rock: :scissors, 
    scissors: :paper, 
    paper: :rock
  }

  # this hash is here to bypass restriction on using conditional operators
  # without it, the code would probably look like 
  #   if condition
  #     p1
  #   else
  #     p2
  #   end
  answers = {
    true => p1, 
    false => p2
  }

  # given the choice of first player, which element can he beat?
  element_dominated_by_first_player = wins[p1]

  # did the second player select the element that first player can beat?
  did_player_one_win = element_dominated_by_first_player == p2

  # pick a winner from our answers hash
  answers[did_player_one_win]
end

winner(:rock, :scissors) # => :rock
winner(:rock, :paper) # => :paper
winner(:scissors, :paper) # => :scissors
于 2012-10-09T09:33:50.570 回答
1

如您所见,这是一个剪刀石头布游戏。def关键字开始一个方法定义。并且end意味着方法的结束。

方法主体的第一行wins = {rock: :scissors, scissors: :paper, paper: :rock}定义了一个名为wins. 这是 ruby​​ 中的语法糖。您也可以将此行写入wins = { :rock => :scissors, :scissors => :paper, :paper => :rock}.

以 a 开头的名称:在 ruby​​ 中称为 Symbol。Symbol 对象表示 Ruby 解释器中的常量名称和一些字符串。

第二行的第一部分{true => p1, false => p2}也是一个哈希。并且wins[p1] == p2可以根据第一行计算出的值。例如,如果您使用winner(:paper, :rock), wins[p1]is :rocknow 和wins[p1] == p2should be调用此方法true{true => p1, false => p2}[true]也是如此p1

ruby 中方法的返回值是最后一个表达式的值。

于 2012-10-09T09:36:34.647 回答