Question

我正在使用python中基于文本的简单刽子手游戏。我有一个粗略的程序运行，但我遇到了两个小问题：

为什么欢迎信息会打印两次？
当用户连续两次输入不在单词中的字母时，第二次，"nope, try again" - 消息不显示，并且不显示当前单词。第一次输入错误的字母时，它可以正常工作。为什么不能第二次运作？

from random import randrange

class HangmanApp:

    def __init__(self, interface):

        self.infile = open("hangman.txt", "r")
        self.interface = textInterface()

        for line in self.infile:
            self.wordlist = line.split()
        self.secretword = self.wordlist[randrange(len(self.wordlist))]

        self.letter_list = list(self.secretword)



    #tests the user's guess

    def guess(self):

        self.new_list = ["_"]*len(self.secretword)

        #keep loop going as long as there are letters in the list
        while self.letter_list != ["_"]*len(self.letter_list):                

            self.guess = self.interface.guess()

            if self.guess in self.letter_list:

                while self.guess in self.letter_list:

                    i = self.letter_list.index(self.guess)

                    #replace that letter from the list with "_" 
                    self.letter_list[self.letter_list.index(self.guess)] = "_"

                    #add the chosen letter to a new list for display
                    self.new_list[i] = self.guess                                   

                    #print list with letter added
                    self.interface.correct(self.new_list)                           

            else:

                self.interface.incorrect(self.new_list)

                self.guess = self.interface.guess()

class textInterface:

    def __init__(self):
        print("Welcome to Hangman!")

    def guess(self):
        guess = input("Guess a letter! ")
        return guess

    def display (self, word):
        string = ' '.join(word)
        print(string)

    def incorrect(self, word):
        print("Nope, try again")
        self.display(word)

    def correct(self, word):
        print("Correct")
        self.display(word)


    def main():
        inter = textInterface()
        app = HangmanApp(inter)
        app.guess()

Answer 1

欢迎讯息会打印两次，因为您要创建两个textInterface个实例：一个在main()中，另一个在HangmanApp.__init__()内。我想你的意思是：

self.interface = interface

而不是

self.interface = textInterface()

在HangmanApp.guess()内，收到错误的猜测（到达else:子句）后，您还有一个额外的guess提示，不需要提示 - 一个不通过您的提示检查代码。我认为这可能导致问题，第二次没有工作。

python中基于简单文本的hangman

1 个答案: