如何糾正我在石頭剪刀布游戲中的錯誤

Question

我在弄清楚如何定義搖滾和確定贏家時遇到了問題，我希望有人可以檢查我所擁有的並告訴我我做錯了什么。 有時，當我將它放入 def main(): 時，我會運行程序，但它會使其成為垃圾郵件，而不是點擊 yes 來玩游戲。 也想知道我的任何縮進是否錯誤，這是否可能是手頭的問題。

import random
def process_computer_choice(): 

    return random.randint(1,3)

def process_player_choice():

    pass
def main():

    def computer_choice():
        return random.choice([1,2])
play_again = 'y'
number_of_tied_games = 0
number_of_player_games = 0
number_of_computer_games = 0
print("Let's play the game of rock, Paper, Scissors.")
while play_again == 'y' or play_again == 'Y':
        computer_choice=process_computer_choice()
        player_choice=process_player_choice()

        if computer_choice == 1: 
            print('the computer chooses rock.')
        elif computer_choice == 2: 
            print('the computer chooses paper.')
        else: 
            print('the computer chooses scissors.')
        #display player choice
        if player_choice == 1: 
            print('You choose rock.')
        elif player_choice == 2:
            print('You choose paper.')
        else: 
            print('You choose scissors.')

            result = determine_winner(player_choice, computer_choice)
if results == 'computer':
            number_of_computer_games += 1
elif result == 'player':
            number_of_player_games += 1
else: 
            number_of_tied_games += 1
    
play_again = input('Do you want to play again? (Enter y for yes): ')
print('There were', number_of_tied_games, 'tie games played.')
print('The computer won', number_of_computer_games, 'game(s).')
print('You won', number_of_player_games, 'game(s).')
def process_computer_choice():  
    choice1 = random.randint(1, 3)


def process_player_choice():
        print('What is your choice? Enter 1 for rock, 2 for paper, or 3 for scissors: ',)
        choice2 = int(input())
        while choice2 != 1 and choice2 !=2 and choice2 != 3:
            print('ERROR: the choice can only be 1, 2 or 3.')
            choice2 = int(input("Please enter a correct choice:"))
        return choice2

def determine_winner(player_choice, computer_choice):
        if coputer_choice == 1:
            if player_choice == 2:
                print('Paper covers rock. You wins!')
                winner = 'player'
        elif player_choice == 3:
             print('Rock crushes scissors. The computer wins!')
             winner = 'computer'
        else: 
             print('The game is tied. Try again.')
             winner = 'tied'
        if computer_choice == 2:
            if player_choice == 1:
                  print('Paper covers rock. The computer wins!')
                  winner = 'computer'
        elif player_choice == 3:
                print(' Scissors cuts paper.You win!')
                winner = 'player'
        else: 
                print('The game is tied. Try again.')
                winner = 'tied'
        if computer_choice == 3:
            if player_choice == 1:
                print('Rock smashes scissors. You win!')
                winner = 'player'
        elif player_choice == 2:
                print('Scissors cuts paper. The computer wins!')
                winner = 'computer'
        else: 
                print('The game is tied. Try again.')
                winner = 'tied'
        return winner
main()

Answer 1

play_again = 'y'    # <- here's the issue
number_of_tied_games = 0
number_of_player_games = 0
number_of_computer_games = 0
print("Let's play the game of rock, Paper, Scissors.")
while play_again == 'y' or play_again == 'Y':    # <-and here

您必須了解計算機程序是自上而下的。 這意味着如果你把這個特定的代碼放在main() （或任何一般的函數main()之外，會發生這種情況：

本質上，您將變量play_again賦值為 'y'，然后計算機繼續檢查play_again是否仍然是 y，它是。 您可以將其視為計算機在“調用”之前完全忽略該函數，這意味着您執行在文件“main()”末尾所做的事情。