如何在不使用最流行的函數的情況下跳出我的 while 循環？（sys.exit、break 等）

Question

我正在制作一個生成隨機數的程序，並要求您猜測 1-100 范圍之外的數字。 輸入數字后，它會根據該數字生成響應。 在這種情況下，如果輸入為 0，則Too high 、 Too low 、 Correct或Quit too soon結束程序（簡化，但基本相同）。

它根據您必須輸入 function 的次數來計算嘗試次數，並使用 while 循環不斷詢問數字，直到您輸入正確為止。 我面臨的問題是，一旦猜測等於隨機數或0 ，我就必須讓它跳出 while 循環。 這通常不是問題，因為您可以使用sys.exit()或其他一些 function，但根據說明我不能使用break 、 quit 、 exit 、 sys.exit或continue 。 問題是我找到的大多數解決方案都是為了打破 while 循環實現break 、 sys.exit或類似的東西，但我不能使用它們。 不過，我使用sys.exit()作為占位符，以便它運行代碼的 rest，但現在我需要找出一種不使用它來中斷循環的方法。 這是我的代碼：

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()

如果有人對如何解決此問題有任何建議或解決方案，請告訴我。

Answer 1

嘗試對您的代碼實施此解決方案：

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，它確實會跳出循環，因為只有在 is_playing 為 True 時循環才會繼續。 這樣，當猜測為 0（您的簡單退出方式）或猜對數字時，您將跳出循環。 希望有所幫助。

我不是全局變量的粉絲，但這是您實現我的解決方案的代碼：

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.'