Python 当用户输入的参数超过函数所能处理的数量时，如何获取与TypeError相关的异常

我试图在代码中实现一个try/exception，当用户输入的参数超过函数预期时，我希望该异常打印“Error”

问题是，我下面的代码没有打印“错误”消息，但它打印的是标准消息：

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: text_analyzer() takes from 0 to 1 positional arguments but 2 were given

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我发现我应该在函数之外使用try/exception来工作

我发现一个更好的解决方案是使用*args。这对我有用

以下是修复代码：

import string


def text_analyzer(text=None, *args):
    ucase = 0
    lcase = 0
    punc = 0
    space = 0

    if len(args) > 0:
        print("Error")
        return
    while not text:
        text = input("What is the text to analyze?")
    for letter in text:
        if letter in string.ascii_uppercase:
            ucase += 1
        elif letter in string.ascii_lowercase:
            lcase += 1
        elif letter in string.punctuation:
            punc += 1
        elif letter in string.whitespace:
            space += 1
    size = len(text)
    print(
        f"The text contains {size} characteres:\n"
        f"- {ucase} upper letters\n"
        f"- {lcase} lower letters\n"
        f"- {punc} punctuation marks\n"
        f"- {space} spaces\n"
    )

---

您尚未显示如何调用

text\u analyzer

；这种错误发生在您进入函数之前。（参数数量错误不是类型错误。）

import string


def text_analyzer(text=None, *args):
    ucase = 0
    lcase = 0
    punc = 0
    space = 0

    if len(args) > 0:
        print("Error")
        return
    while not text:
        text = input("What is the text to analyze?")
    for letter in text:
        if letter in string.ascii_uppercase:
            ucase += 1
        elif letter in string.ascii_lowercase:
            lcase += 1
        elif letter in string.punctuation:
            punc += 1
        elif letter in string.whitespace:
            space += 1
    size = len(text)
    print(
        f"The text contains {size} characteres:\n"
        f"- {ucase} upper letters\n"
        f"- {lcase} lower letters\n"
        f"- {punc} punctuation marks\n"
        f"- {space} spaces\n"
    )