Question

好的，我正在为一个项目上课，我坚持这个逻辑错误。我在我的代码中修复了原始错误之后的其他原因......我觉得我知道逻辑错误在哪里。我只是不确定如何解决它。我可以得到一些帮助吗？

提前致谢。

import random

def display_title():
    print("Guess the number!")
    print()

def set_limit():
    print("Enter the upper limit for the range of numbers: ")
    limit = int(input())
    return limit

def count(): ## had to add count being defined as below it was unrecognized by python. 
    count +=1

def play_game(limit):
    global count
    number = random.randint(1, limit)
    print("I'm thinking of a number from 1 to " + str(limit) + "\n")
    while True:
        guess = int(input("Your guess: "))
        if guess < number:
            print("Too low. ")
            count ## See def count
        elif guess >= number:
            print("Too high. ")
            count ## See def count
        elif guess == number: ## Pretty sure my logic error is here <---- 
            print("You guessed it in " + str(count) + " tries.\n")
            return 

def main(): ## syntax error, no : was here
    display_title()
    again = "y"
    while again.lower() == "y":
        limit = set_limit()
        play_game(limit) ## limit wasn't set inside, causing a missing positional argument 
        again = input("Play again? (y/n): ")
        print()
    print("Bye!")

if __name__ == "__main__":
    main()

Answer 1

我有个主意。尝试将上面的＆gt; =符号3行更改为＆gt;数。因为否则它永远不会到达＆＃34;如果猜测==数字＆＃34;因为它会停在＆＃34;如果猜测＆gt; =数字＆＃34;每次即使猜测是正确的。

这是因为＆gt; =表示大于或等于。我真的希望这会有所帮助。

Answer 2

您无需使用print()打印空行。
您可以使用input提示用户之前不需要单独的print声明。
count可以在play_game内定义，并且应该初始化为1而不是默认0，因为即使用户在第一次尝试时猜到他们也有1次猜测
elif guess >= number:需要更改为elif guess > number:
如果answer的第一个字符转换为小写等于y，您可以再次调用main()重新启动游戏。

我在下面的代码中解决了上述问题：

import random

def display_title():
  print("Guess the number!\n")

def set_limit():
  limit = int(input("Enter the upper limit for the range of numbers: "))
  return limit

def play_game(limit):
  count = 1
  number = random.randint(1, limit)
  print("I'm thinking of a number from 1 to " + str(limit) + "\n")
  while True:
    guess = int(input("Your guess: "))
    if guess < number:
      print("Too low. ")
      count += 1
    elif guess > number:
      print("Too high. ")
      count += 1
    elif guess == number: 
      print("You guessed it in " + str(count) + " tries.\n")
      break

def main(): 
  display_title()
  limit = set_limit()
  play_game(limit)
  again = input("Play again? (y/n): ")
  if again[0].lower() == 'y':
    main()
  else:
    print("Bye!")

if __name__ == "__main__":
  main()

试试here!

python 3逻辑错误让我搞砸了

2 个答案: