python - 如果不立即重新引发异常回溯，则隐藏

Question

我有一段类似这样的代码:

import sys

def func1():
    func2()

def func2():
    raise Exception('test error')

def main():
    err = None

    try:
        func1()
    except:
        err = sys.exc_info()[1]
        pass

    # some extra processing, involving checking err details (if err is not None)

    # need to re-raise err so caller can do its own handling
    if err:
        raise err

if __name__ == '__main__':
    main()

当 func2 引发异常时，我收到以下回溯:

Traceback (most recent call last):
  File "err_test.py", line 25, in <module>
    main()
  File "err_test.py", line 22, in main
    raise err
Exception: test error

从这里我看不出异常来自哪里。原始回溯丢失。

如何保留原始回溯并重新提出？我想看到类似的东西:

Traceback (most recent call last):
  File "err_test.py", line 26, in <module>
    main()
  File "err_test.py", line 13, in main
    func1()
  File "err_test.py", line 4, in func1
    func2()
  File "err_test.py", line 7, in func2
    raise Exception('test error')
Exception: test error

Answer 1

空白的raise引发最后一个异常。

# need to re-raise err so caller can do its own handling
if err:
    raise

如果你使用 raise something Python 无法知道 something 是刚刚捕获的异常，还是带有新堆栈跟踪的新异常。这就是保留堆栈跟踪的空白 raise 的原因。

Reference here