如果不立即引发异常回溯，则将其隐藏

Question

我有一段类似于以下的代码：

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，我会收到以下回溯：

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

从这里我看不到异常的来源。原始回溯丢失。

如何保留原始回溯并重新提出？我想看类似的东西：

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

如果您使用raise somethingPython，则无法知道something是之前捕获的异常还是带有新堆栈跟踪的新异常。这就是为什么有空白raise保留堆栈跟踪的原因。

这里参考