首页Traceback (most recent call last): File "D:\zjj\main.py", line 17, in <module> a=encrypt(19,25) ^^^^^^^^^^^^^^ File "D:\zjj\main.py", line 15, in encrypt c=c+ci ^ UnboundLocalError: cannot access local variable 'c' where it is not associated with a value

Traceback (most recent call last): File "D:\zjj\main.py", line 17, in <module> a=encrypt(19,25) ^^^^^^^^^^^^^^ File "D:\zjj\main.py", line 15, in encrypt c=c+ci ^ UnboundLocalError: cannot access local variable 'c' where it is not associated with a value

时间: 2023-11-13 10:14:41 浏览: 125

浅谈Python traceback的优雅处理

刚接触Python的时候，简单的异常处理已经可以帮助我们解决大多数问题，但是随着逐渐地深入，我们会发现有很多情况下简单的异常处理已经无法解决问题了，如下代码，单纯的打印异常所能提供的信息会非常有限。 def func1(): raise Exception("--func1 exception--") def main(): try: func1() except Exception as e: print e if __name__ == '__main__': main() 执行后输出如下： –func1 exception– 通过示例，我们发现

这是一个 Python 报错信息，意思是在执行代码时出现了错误。这个错误是因为在 encrypt 函数中，变量 c 在使用前没有被赋初值。可以在函数内部添加 `c = ""` 来初始化变量 c，解决这个问题。下面是修正后的代码： ```python def encrypt(k, m): c = "" for i in m: if i.isalpha(): ci = chr((ord(i.lower())-97+k)%26+97) else: ci = i c = c+ci return c ```

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Traceback (most recent call last): File "D:\zjj\main.py", line 17, in <module> a=encrypt(19,25) ^^^^^^^^^^^^^^ File "D:\zjj\main.py", line 15, in encrypt c=c+ci ^ UnboundLocalError: cannot access local variable 'c' where it is not associated with a value

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Traceback (most recent call last): File "greeter.py", line 1, in <module> name = input("Please enter your name: ") File "<string>", line 1, in <module> NameError: name 'yue' is not defined解释这个错误

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