为什么我的一个函数在不需要的时候只是循环？

Question

我试图创建一个简单的代码猜测游戏，用户可以选择随机生成的代码的最小和最大数量。用户必须尝试并猜测代码才能获胜。当我运行我的代码时，get_range（）函数工作，然后它继续执行get_guess（）函数。但是，当用户输入他/她的输入进行猜测时，代码循环回到get_range（）函数的开头。请问有人能帮忙吗？提前谢谢。代码：

import random
import string

print("Welcome to Code Crunchers!")


def get_range():
    Min = str(input("Enter the minimum number that the code can be: "))
    Max = str(input("Enter the maximum number that the code can be: "))

    Check_Min = Min.isdigit()
    Check_Max = Max.isdigit()

    if Check_Min != True or Check_Max != True:
        print("Input must only contain integers!")
        get_range()
    elif Min == Max:
        print("Minimum and maximum number must not be equivalent!")
        get_range()
    elif Min > Max:
        print("Maximum number must be greater than minimum number!")
        get_range()
    else:

        Random_Number = random.randrange(int(Min), int(Max))

        get_guess()

        return Random_Number


def get_guess():
    Guess = str(input("Enter your guess: "))

    Check_Guess = Guess.isdigit()

    if Check_Guess != True:
        print("Input must only contain integers!")
        get_guess()
    else:
        validate()

    return Guess


def validate():
    Random_Number = get_range()
    Tries = locals()
    Guess = get_guess()
    Length = len(str(Random_Number))
    Digits_Correct = 0

    if Guess == Random_Number:
        print("Well done! You guessed the number in", Tries, " tries!")
    else:
        Digits = ["?"] * Length

        Tries += 1

        for i in range(0, int(Length)):
            if Guess[i] == Random_Number[i]:
                Digits[i] = Guess[i]
                Digits_Correct += 1
            else:
                continue

        if int(Length) > Digits_Correct > 0:
            print("Not quite! You got", Digits_Correct, " digits correct.")
            print(Digits)
            get_guess()
        elif Digits_Correct == 0:
            print("None of your digits match!")
            get_guess()


def play_again():
    Choice = input("Do you want to play again (y/n)?")

    if Choice != "y" or Choice != "n" or Choice != "Y" or Choice != "N":
        print("Please choose a valid option!")
        play_again()
    elif Choice == "y" or Choice == "Y":
        get_range()
    elif Choice == "n" or Choice == "N":
        exit()


get_range()

Answer 1

请看下面的代码：

def get_range():
    ... 
    else:
        ...

        get_guess()

        return Random_Number

def get_guess():
    ...
    else:
        validate()

    return Guess

def validate():
    Random_Number = get_range()
    Tries = locals()
    Guess = get_guess()
    ...

假设您在get_guess并且到达else关闭点，那么您调用validate。下面是发生的情况：

1. get_guess调用validate
2. validate立即调用get_range
3. get_range调用get_guess
4. 现在我们回到get_guess，请参见（1）

因此，您的代码进入无限间接递归

请注意，它在validate中永远不会通过Random_Number = get_range()，您在get_range和validate中都调用了get_guess

因此，在将随机数返回到Random_Number = get_range()之前，get_range将尝试get_guess并立即丢弃其返回值（这就是get_guess()的作用）。假设get_range最终返回。现在您将再次调用Guess = get_guess()，从而要求用户猜两次。我认为这里有一个逻辑缺陷