Python二等分數字猜謎游戲

Question

我正在嘗試編寫一個簡單的二等分方法問題，只要我沒有注釋掉的條件語句，它就可以很好地工作。 這是什么原因呢？ 這不是作業問題。

low = 0 
high = 100 
ans = (low+high)/2 
print "Please think of a number between 0 and 100!"
print "Is your secret number " + str(ans) + "?"
response = raw_input("Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly. ") 
response = str(response)
while response != "c": 
   if response == "h":
        high = ans 
        ans = (low + high)/2 
        print "Is your secret number " + str(ans) + "?"
        response = raw_input("Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly. ")  
        response = str(response) 

   if response == "l": 
        low = ans 
        ans = (low + high)/2 
        print "Is your secret number " + str(ans) + "?"
        response = raw_input("Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly. ")  
        response = str(response)

   if response == "c" :
        break

  # if response != "c" or response != "h" or response != "l":
   #     response = raw_input("Please enter a 'h', 'l', or 'c' ")
    #    response = str(response)

print "Game over. Your secret number was: " + str(ans)

這是因為while循環與while循環具有相同的條件嗎？ 如果是這樣，改變這一狀況的最佳方法是什么？

Answer 1

該條件將永遠是正確的，因為您正在將不平等與多個事物進行比較。 這就像問“如果此字符不是c ， 還是不是h ， 還是不是l ， 請執行此操作 。一次不能是三件事，因此它將始終評估為true 。

相反，您應該if response not in ['c','h','l']使用if response not in ['c','h','l'] ，這基本上就像在上面的句子中替換or或 with 和 。 甚至對您而言更好，只需使用else語句，因為您的現有條件已經可以確保您嘗試檢查的內容。

Answer 2

我建議您使用double while循環：

while True:
    response = None
    print "Is your secret number " + str(ans) + "?"
    response = str(raw_input("Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly."))

    while response not in ['h', 'c', 'l']:
        response = str(raw_input("Please enter a 'h', 'l', or 'c' "))

    if response == 'c':
         break

    if response == 'l':
        low = ans 
    else:
        high = ans 
    ans = (low + high)/2