為什么while循環在if循環中不起作用？

Question

    def n():
        name = input('What is the missing animal?')
        if name == 'dog':
                print('Well done')
        else:
         print('Sorry this is not right')
         rep= 0 
         while rep < 5:
                  n()
         rep = rep + 1
         if rep == 5:
              print ('You have guessed incorrectly 5 times.)

當我運行此程序並得到錯誤的答案時，程序將不斷重復，而不是重復最多5次。

有任何想法嗎？

Answer 1

多么尷尬的遞歸。 :)

問題在於rep變量是局部作用域的，即未傳遞給遞歸調用。

你應該把while外面用success與變量while為了測試是否需要再次循環。

無需遞歸。

編輯：這樣：

def n():
    rep= 0 
    success = 0
    while rep < 5 or success == 1:
        name = input('What is the missing animal?')
        if name == 'dog':
            success = 1
        else:
            print('Sorry this is not right')
            rep = rep + 1
    if rep == 5:
        print ('You have guessed incorrectly 5 times.')
    elif success == 1:
        print('Well done')

抱歉縮進。

Answer 2

def n():
    for rep in range(5):
        name = input('What is the missing animal?')
        if name == 'dog':
            print('Well done')
            break
        else:
            print('Sorry this is not right')
    else:
        print ('You have guessed incorrectly 5 times.')

因為您知道要循環幾次，所以for （也許）更合適。 for循環的else子句可以處理您沒有得到正確答案的情況。

Answer 3

您一直在else語句中反復調用n（）方法。 我相信這段代碼將滿足您的需求：

def n():
    rep= 0
    while rep < 5:
        name = input('What is the missing animal? ')
        if name == 'dog':
            print('Well done')
            break
        else:
            print('Sorry this is not right')
        rep = rep + 1
    if rep >= 5:
        print ('You have guessed incorrectly 5 times.')

除非您得到正確的答案，否則此循環將運行5次。 如果答案正確，則循環將break ，這意味着它將停止運行。 最后，它檢查rep是否大於（永遠不會）或等於（在第5次循環中發生），並且如果循環了5次，則打印結束消息。

Answer 4

這是遞歸的正確方法。 盡管這是遞歸的尾部，但我自己將其展開為類似@Prune的循環。

def n(rep=0):
    if n >= 5:
        print ('You have guessed incorrectly 5 times.')
    else:
        name = input('What is the missing animal?')
        if name == 'dog':
            print('Well done')
        else:
            n(rep+1)