Question

我无法为这个确切的问题找到解决方案，以至于我希望有人可以对此有所了解。

我在这里有此代码：

def numberOfPorts():
    try:
        ports = int(input("How many ports do you want to configure? "))
    except:
        print("You did not specify a number.")
        numberOfPorts()
    return ports

现在，如果我运行此代码并输入数字，则该函数运行正常。

numberOfPorts()
How many ports do you want to configure? 10
10

但是当我第二次运行此函数并指定一个字符串时，将触发除外代码，并执行我的print语句，并且输入函数再次要求输入。这就是我要的。然后，我给脚本一个数字，但随后出现以下错误：

numberOfPorts()
How many ports do you want to configure? foobar
You did not specify a number.
How many ports do you want to configure? 10
Traceback (most recent call last):
  Python Shell, prompt 53, line 1
  Python Shell, prompt 49, line 8
builtins.UnboundLocalError: local variable 'ports' referenced before assignment

确保这真的很简单，我只需要对根本原因有一点了解。

Answer 1

您可以执行以下操作-

def numberOfPorts():
    try:
        ports = int(input("How many ports do you want to configure? "))
        return ports
    except:
        print("You did not specify a number.")
        return numberOfPorts()

但是对于这个问题，递归是过大的。为什么？因为我们只会用函数调用填充内存堆栈，直到获得整数。因此while循环就足够了-

def numberOfPorts():
    ports = None
    while ports is None:
        try:
            ports = int(raw_input("How many ports do you want to configure? "))
        except ValueError:
            print("You did not specify a number.")
    return ports

Builtins.UnboundLocalError：局部变量

1 个答案: