Python UnboundLocalError求助：变量number赋值前被引用

嘿，这个问题我见过不少编程新手碰到，咱们来拆解一下到底咋回事：

问题根源 #

当你第一次输入非数字（比如hkjadhjkas）时，try块里的number = int(input("--> "))执行失败，直接跳到except块：

• 打印"Not a number!"
• 调用straight()递归，让用户重新输入

等这次递归执行完成（比如你输入20成功通过），程序会回到上一层的straight()函数，继续执行后面的代码：if number % 2 == 0:——但这一层的number变量根本没被赋值！因为刚才的try块失败了，number从未被定义，所以就抛出了UnboundLocalError。

而第一次直接输入数字时，try块成功给number赋值，后面的判断能正常执行，所以不会报错。

修复方案 #

方案1：快速修复递归逻辑

最直接的修复是，在except块调用递归后，加上return，让函数直接终止，不再执行后面未定义number的代码：

def straight(): 
    print("You went straight ahead") 
    print("You run into a locked door. You try to break down the door but it doesn't work..") 
    print("Then you see the writing on the wall next to the door..") 
    print("Call a number and then we call it even!") 
    try: 
        number = int(input("--> ")) 
    except: 
        print("Not a number!")
        straight()
        return  # 加这一行，避免执行后面未定义number的代码
    if number % 2 == 0: 
        print(f"You cracked the code! Your even number {number} gave you access to this door") 
        print("The door opens..") 
        final_room() 
    elif number % 2 >= 1: 
        dead(f"The wrong number ({number}) caused the tunnel to collapse!") 
    else: 
        print("test else?") 
        straight() 

方案2：用循环代替递归（更推荐）

递归如果遇到用户多次输入错误，可能会触发函数调用栈溢出，用循环处理输入验证更安全：

def straight(): 
    print("You went straight ahead") 
    print("You run into a locked door. You try to break down the door but it doesn't work..") 
    print("Then you see the writing on the wall next to the door..") 
    print("Call a number and then we call it even!") 
    
    # 循环直到用户输入有效的数字
    while True:
        try: 
            number = int(input("--> ")) 
            break  # 输入正确，跳出循环
        except: 
            print("Not a number! Please try again.")
    
    if number % 2 == 0: 
        print(f"You cracked the code! Your even number {number} gave you access to this door") 
        print("The door opens..") 
        final_room() 
    else: 
        dead(f"The wrong number ({number}) caused the tunnel to collapse!") 

（注：这里把elif number %2 >=1改成了else，因为整数要么是偶数要么是奇数，没必要额外判断）

内容的提问来源于stack exchange，提问作者Marije