Python 对分猜数游戏

Question

我正在尝试编写一个简单的二分法问题，只要我没有已注释掉的特定条件语句，它就可以很好地工作。这是什么原因？这不是作业题。

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)

这是因为while循环和while循环的条件一样吗？如果是这样，改变这种情况的最佳方法是什么？

Answer 1

该条件将始终为真，因为您正在比较多个事物的不平等性。这就像问“如果这个字符不是 c, or 如果它不是 h, or 如果它不是 l，做这个。它不能同时是三个东西，所以它总是评估为真。

相反，您应该使用 if response not in ['c','h','l']，这基本上就像用 替换 or和 在上面的句子中。或者在您的情况下更好，只需使用 else 语句，因为您现有的条件已经确保您要检查的内容。