Python 3：猜數字 – 回顧 · JavaBeginnersTutorial 中文系列教程

# Python 3：猜數字 – 回顧 > 原文： [https://javabeginnerstutorial.com/python-tutorial/python-3-guess-the-number-the-return/](https://javabeginnerstutorial.com/python-tutorial/python-3-guess-the-number-the-return/) 我們之前已經看過這個示例。在那里，我包括了很多處理用戶輸入和主要邏輯的循環……現在該重構應用以使用函數了。這應該使代碼易于閱讀，以后我們可以重用部分代碼。這些可重復使用的部分之一是數字讀數。我們有兩個要點，我們希望用戶輸入數字并編寫了略微相同的代碼。這是開始重構并將其提取為一個函數的好地方。因此，讓我們將其提取為一個函數。唯一的區別是我們打印出的詢問用戶輸入數字的消息。可以將其作為函數的參數來處理，因為這是唯一的可變部分。因此，讓我們看一下函數的定義： ```py def ask_user_for_number(message_text): while True: try: return int(input(message_text)) except ValueError: print("This was not a number!") continue ``` 如您所見，這是一種非常簡單的方法，我們要求用戶輸入內容，如果輸入的是數字，則將其返回。自然，我們無法使用此方法處理將數字驗證為 100 或 1000 的情況，因此我們也必須在那里修改代碼塊。在此簡單修改的??最后，我們為應用提供了以下代碼（并且具有相同的功能）： ```py __author__ = 'GHajba' import random def ask_user_for_number(message_text): while True: try: return int(input(message_text)) except ValueError: print("This was not a number!") continue return number while True: max_number = 0 while max_number != 100 and max_number != 1000: max_number = ask_user_for_number('Should the secret number between 1 and 100 or 1 and 1000? ') if max_number == 100: guess_count = 7 else: guess_count = 10 print('You have chosen {}, you will have {} guesses to find the secret number.'.format(max_number, guess_count)) secret_number = random.randint(1, max_number) print('I have chosen the secret number...') guesses = 0 while guess_count - guesses: guesses += 1 guessed = ask_user_for_number("What's your guess? ") if guessed == secret_number: print('Congrats, you have Won!') break elif guessed > secret_number: print('The secret number is lower...') else: print('The secret number is higher...') else: print("Sorry, you lose.") print("The secret number was ", secret_number) answer = '' while answer.lower() not in ['yes', 'no', 'y', 'n']: answer = input("Do you want to play another round? (yes / no) ") if 'no' == answer or 'n' == answer: break ``` 看起來很不錯，沒有重復的代碼，而且非常精簡。但是，如果您對測試有所了解，您可能會說這段代碼有一個很大的“主”塊，很難測試。我必須同意。單元測試（盡管我將在下一章中進行介紹）在這里會很麻煩。解決方案是將這個較大的主循環拆分為較小的函數，這些函數可以單獨進行測試。例如，求值用戶是贏還是輸。為此，我們可以編寫一個函數，該函數將秘密數字，猜測，猜測計數和猜測數字作為輸入，并對消息進行求值以告知用戶。但是一個函數的四個參數很多，因此現在讓我們對其進行劃分。基于這些輸入，我將創建一個函數來告訴用戶秘密數字是猜中的數字是更高還是更低。 ```py def user_won(guessed, secret): if guessed == secret_number: print('Congrats, you have Won!') return True if guessed > secret_number: print('The secret number is lower...') else: print('The secret number is higher...') ``` 在`user_won`函數中，我們使用`return`語句指示用戶是否贏了。如果不是，則返回隱式`None`，其結果為`false`。我們可以測試的另一件事是詢問用戶是否要再玩一輪。 ```py def want_continue(): answer = '' while answer.lower() not in ['yes', 'no', 'y', 'n']: answer = input("Do you want to play another round? (yes / no) ") return answer in ['yes','y'] ``` 完成所有這些更改后，讓我再次向您顯示完整代碼： ```py __author__ = 'GHajba' import random def ask_user_for_number(message_text): while True: try: return int(input(message_text)) except ValueError: print("This was not a number!") continue return number def user_won(guessed, secret): if guessed == secret_number: print('Congrats, you have Won!') return True if guessed > secret_number: print('The secret number is lower...') else: print('The secret number is higher...') def want_continue(): answer = '' while answer.lower() not in ['yes', 'no', 'y', 'n']: answer = input("Do you want to play another round? (yes / no) ") return answer in ['yes','y'] while True: max_number = 0 while max_number != 100 and max_number != 1000: max_number = ask_user_for_number('Should the secret number between 1 and 100 or 1 and 1000? ') if max_number == 100: guess_count = 7 else: guess_count = 10 print('You have chosen {}, you will have {} guesses to find the secret number.'.format(max_number, guess_count)) secret_number = random.randint(1, max_number) print('I have chosen the secret number...') guesses = 0 while guess_count - guesses: guesses += 1 guessed = ask_user_for_number("What's your guess? ") if user_won(guessed, secret_number): break else: print("Sorry, you lose.") print("The secret number was ", secret_number) if not want_continue(): break ```