Ruby - 秘密号码游戏，如何让游戏重启？

Question

我试图让我的秘密号码游戏正常运作。

我遇到了让游戏重启的问题：

你没有更多的猜测了 你赢了比赛

我使用了while循环，但是通过函数/方法访问它有麻烦。

编辑：

所以基本上玩家有3次尝试猜测秘密号码。如果他们赢了，他们会被问到是否要重新开始比赛。如果他们输了，他们也会被问到是否要重试。

这是代码：

先谢谢你们

 def get_input
        gets.chomp
    end

    puts "Welcome to the Secret Number Game! Please tell me your name"
    player_name = get_input
    puts "Welcome #{player_name}!"
    puts "Guess a number between 1 - 10. You only have 3 attempts!"
    secret_number = 1 + rand(10)
    restart = true

    while restart

        def playAgain ( restart )
            puts "Would you like to play again? (y/n)"
            answer = get_input
            if answer == "n"
                restart = false
            end
        end

    def guess_check ( player_input, secret_number )     
        if player_input > secret_number
            puts "Wrong, sorry! Too high!"
        elsif player_input < secret_number
            puts "Wrong, sorry! Too low!" 
        else
            puts "Congratulations, you've guessed the secret number! #{[secret_number]}"
            playAgain ( restart )
        end
    end


        ############################## START GAME ###########################################

        guesses = []
        attempts = 3    

        while attempts

            attempts = attempts - 1
            if attempts == -1
                puts "Sorry, you have no more tries"
                playAgain ( restart )
            else
                puts "Guess the secret number (You have #{attempts} tries left):"
            end

            player_input = get_input.to_i
            guesses.push ( player_input )
            guess_check( player_input, secret_number )
            puts "You've guessed - #{guesses}"

        end

            playAgain ( restart )

    end

Answer 1

红宝石中的{p> 0是真实的，而不是虚假的，与大多数语言相矛盾。要打破while循环，应该明确检查它是否大于零：

- while attempts
+ while attempts > 0

或更多rubyish：

3.downto(0) do |attempts| 
    ...  
end

<强> UPD

重启的部分。您将获得playAgain中定义的本地重新启动变量。谷歌用于ruby范围，简而言之：函数中的局部变量在这个函数的范围之外是不可见的;参数按值传递给函数。也就是说，在restart = false内定义playAgain根本没有任何意义，因为局部变量在函数范围内死亡。可能的解决方案是声明实例变量@restart。在这种情况下，您不需要将其作为参数传递。但是，完成任务的最合理方法是从playAgain返回一个布尔值：

def playAgain
  puts "Would you like to play again? (y/n)"
  answer = get_input
  answer != "n" # return a boolean from the function in last statement
end

然后整个范围看起来像：

def playAgain
  puts "Would you like to play again? (y/n)"
  answer = get_input
  answer != "n" # return a boolean from the function in last statement
end

def guess_check ( player_input, secret_number )     
  if player_input > secret_number
    puts "Wrong, sorry! Too high!"
  elsif player_input < secret_number
    puts "Wrong, sorry! Too low!" 
  else
    puts "Congratulations! Result: #{secret_number}"
  end
  player_input == secret_number # return true on success
end

# let’s go!
loop do # main loop
  guesses = []
  3.downto(1) |attempts| # 3 attemts
    puts "Guess the secret number (You have #{attempts} tries left):"

    player_input = get_input.to_i
    guesses.push(player_input)
    # you have to return true/false from guess check 
    #     and rerun the topmost loop if success
    # see the trick with logical and below:
    #     on normal execution the loop will return it’s initial
    #     (3 in this particular case)
    #     on successful guess is will break, returning false  
    break false if guess_check(player_input, secret_number)
    puts "You've guessed - #{guesses}"
  end && puts "Sorry, you have no more tries"

  break unless playAgain # will break a topmost loop unless `y` input
end