在Mastermind Gam中重复函数的问题

Question

我正在设计一个用python玩的策划游戏。但是，当我试图设置一个函数，使其在尝试不完全正确的情况下重复时，我遇到了一些问题。你知道吗

我的代码分为两部分。对于第一部分，它要求用户输入正确的号码，然后第二个用户尝试输入他的尝试号码。代码的第二部分将他的尝试分解为数字列表，并计算正确整数的数量和正确位置的整数的数量，然后如果答案不完全正确，程序会要求用户进行第二次输入。你知道吗

def getnumber():
predestine = input("Please input your test number")
a = str(predestine)
attempt()

def attempt():
  attempt = input("Attempt:")
  b = str(attempt)
  correctrecord = []
  sequencerecord = []
  for i in b:
      if i in a:
          correctrecord.append(1)
  for i in range(0,4):
      if b[i] == a[i]:
        s  equencerecord.append(1)

  correctlength = len(correctrecord)
  sequencelength = len(sequencerecord)

  print(f"You have made {correctlength} correct attempts, and of these {sequencelength} are of correct positions")

  if sequencelength == 4:
      print("You have won, the game is ended")
  else:
      return attempt()

问题在于最后一个代码：return attempt（）。它似乎无法重复该函数，并出现“str object not callable”错误。你知道吗

Answer 1

您可以多次使用相同的变量名。Python函数是一级公民，它允许您执行以下操作：

# define a function by name r
def r():
    return 1

print(type(r))  # <class 'function'>     now r is a function

# reassign name r to be a string
r = "22"
print(type(r))  # <class 'str'>          now r is a string

如果你做了r()，现在你得到了TypeError: 'str' object is not callable

---

您的代码使用全局变量，然后再次调用自己的函数，这可能导致递归溢出—请参见What is the maximum recursion depth in Python, and how to increase it?

当存在重复项时，计算正确“位数”的数量时，您将得到错误的结果-请尝试使用“1122”作为正确值，使用“1234”作为尝试值。你知道吗

递归是不需要你的游戏代码。我对它进行了一些调整，以展示一种不同的方式：

def getnumber(text):
    """Loops until a number is inputted. Returns the number as string.
    'text' is the prompt for the user when asking for a number."""
    while True:
        try:
            predestine = int(input(text).strip())
            return str(predestine)
        except:
            print("Only numbers allowed!")

def attempt(correct_number):
    """One game round, returns True if correct result was found."""
    attempt = getnumber("Attempt: ")

    # avoid double-counting for f.e. 1212  and 1111 inputs
    correct_digits = len(set(attempt) & set(correct_number))

    sequencelength = 0                 # no need to collect into list and use len
    for i,c in enumerate(attempt):     # simply increment directly
        if c == correct_number[i]:
            sequencelength += 1 

    print(f"You have found {correct_digits} correct digits, and of these {sequencelength} are of correct positions.")
    if len(attempt) < len(correct_number):
        print("Your number has too few digits.")
    elif len(attempt) > len(correct_number):
        print("Your number has too many digits.")

    return correct_number == attempt

# game - first get the correct number
number = getnumber("Please input your test number: ")

# loop until the attempt() returns True
ok = False
while not ok:
    ok = attempt(number)

print("You have won, the game is ended")  

输出：

Please input your test number: 1234
Attempt: 1111
You have found 1 correct digits, and of these 1 are of correct positions.
Attempt: 1212
You have found 2 correct digits, and of these 2 are of correct positions.
Attempt: 1321
You have found 3 correct digits, and of these 1 are of correct positions.
Attempt: 1234
You have found 4 correct digits, and of these 4 are of correct positions.
You have won, the game is ended

Answer 2

代码中的问题在于变量阴影。你知道吗

您的重复函数位于一个名为attempt的全局变量中。然后，在attempt函数中定义一个attempt字符串变量，该变量是该函数的局部变量，因此暂时隐藏了保存该函数的全局attempt变量。你知道吗

因此，调用attempt()失败，因为您实际上是在尝试调用字符串。你知道吗

解决方案是将局部字符串变量attempt重命名为不覆盖全局变量：

def attempt():
    attempt_ = input("Attempt:")
    b = str(attempt_)
    correctrecord = []
    sequencerecord = []
    for i in b:
        if i in a:
            correctrecord.append(1)
    for i in range(0,4):
        if b[i] == a[i]:
            sequencerecord.append(1)

    correctlength = len(correctrecord)
    sequencelength = len(sequencerecord)

    print(f"You have made {correctlength} correct attempts, and of these {sequencelength} are of correct positions")

    if sequencelength == 4:
        print("You have won, the game is ended")
    else:
        return attempt()