Python二分号猜测游戏

Question

我试图写一个简单的二分法方法问题,只要我没有一个我已经注释掉的某个条件语句,它就能完美地运行.这是什么原因？这不是一个家庭作业问题.

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循环与while循环具有相同的条件吗？如果是这样,改变这个的最佳方法是什么？

Answer 1

这种情况永远是真实的,因为你正在将不平等与多种事物进行比较.这就像问"如果这个角色不是c,或者它不是h,或者它不是l,那么这样做.它不能同时是三件事,所以总是会评价为真.

相反,你应该使用if response not in ['c','h','l'],这基本上是像更换或用和上面的句子.或者在您的情况下更好,只需使用一个else声明,因为您现有的条件已经确保您要检查的内容.