如何在不使用最流行的函数的情况下跳出我的 while 循环？（sys.exit、break 等）

Question

我正在制作一个生成随机数的程序，并要求您猜测 1-100 范围之外的数字。 输入数字后，它会根据该数字生成响应。 在这种情况下，如果输入为 0，则Too high 、 Too low 、 Correct或Quit too soon结束程序（简化，但基本相同）。

它根据您必须输入 function 的次数来计算尝试次数，并使用 while 循环不断询问数字，直到您输入正确为止。 我面临的问题是，一旦猜测等于随机数或0 ，我就必须让它跳出 while 循环。 这通常不是问题，因为您可以使用sys.exit()或其他一些 function，但根据说明我不能使用break 、 quit 、 exit 、 sys.exit或continue 。 问题是我找到的大多数解决方案都是为了打破 while 循环实现break 、 sys.exit或类似的东西，但我不能使用它们。 不过，我使用sys.exit()作为占位符，以便它运行代码的 rest，但现在我需要找出一种不使用它来中断循环的方法。 这是我的代码：

import random
import sys

def main():
    global attempts
    attempts = 0
    guess(attempts)
    keep_playing(attempts)
    
def guess(attempts):
    number = random.randint(1,100)
    guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
    while guess != 0: 
        if guess != number:
            if guess < number:
                print("Too low, try again")
                attempts += 1
                guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
            elif guess > number:
                print("Too high, try again")
                attempts += 1
                guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
        else:
            print()
            print("Congratulations! You guessed the right number!")
            print("There were", attempts,"attempts")
            print()
            #Ask if they want to play again
            sys.exit()#<---- using sys.exit as a placeholder currently
    else:
        print()
        print("You quit too early")
        print("The number was ",number,sep='')
        #Ask if they want to play again
        sys.exit()#<----- using sys.exit as a placeholder currently
    
    

def keep_playing(attempts):
    keep_playing = 'y'
    if keep_playing == 'y' or keep_playing == 'n':
        if keep_playing == 'y':
            guess(attempts)
            keep_playing = input("Another game (y to continue)? ")
        elif keep_playing == 'n':
            print()
            print("You quit too early")
            print("Number of attempts", attempts)
main()

如果有人对如何解决此问题有任何建议或解决方案，请告诉我。

Answer 1

尝试对您的代码实施此解决方案：

is_playing = True

while is_playing:
    if guess == 0:
        is_playing = False
        your code...
    else:
        if guess == number:
            is_playing = False
            your code...
        else:
            your code...

不使用任何 break 等，它确实会跳出循环，因为只有在 is_playing 为 True 时循环才会继续。 这样，当猜测为 0（您的简单退出方式）或猜对数字时，您将跳出循环。 希望有所帮助。

我不是全局变量的粉丝，但这是您实现我的解决方案的代码：

import random

def main() -> None:
    attempts = 0
    global is_playing
    is_playing = True
    while is_playing:
        guess(attempts)
        keep_playing()
    
def guess(attempts: int) -> None:
    number = random.randint(1,100)
    print(number)
    is_guessing = True
    while is_guessing:
        attempts += 1
        guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
        if guess == 0:
            is_guessing = False
            print("\nYou quit too early.")
            print("The number was ", number,sep='')
        else:
            if guess == number:
                is_guessing = False
                print("\nCongratulations! You guessed the right number!")
                print("There were", attempts, "attempts")
            else:
                if guess < number:
                    print("Too low, try again.")
                elif guess > number:
                    print("Too high, try again.")
                
def keep_playing() -> None:
    keep_playing = input('Do you want to play again? Y/N ')
    if keep_playing.lower() == 'n':
        global is_playing
        is_playing = False

main()

提示：而不是"There were", attempts, "attempts"做： f'There were {attempts} attempts.'