Question

有没有办法防止在将堆栈升级到程序的最高级别时多次捕获异常？

这是一个非常简化的代码示例，用于说明这种现象：

def try_except_block(smthg):
   try:
     smthg.run()
   except Exception as e:
     print("WRONG")
     raise e

def re_call():
   f2 = Foo(True) # will throw an exception
   try_except_block(f2)

class Foo:
  def __init__(self, t):
    self.throw = t
  def run(self):
    if self.throw:
      raise KeyError
    else:
      re_call()

if __name__ == '__main__':
  f = Foo(False) # won't throw an exception
  try_except_block(f)

输出：

WRONG
WRONG
Traceback (most recent call last):
  File "exception_loosing_args.py", line 26, in <module>
    try_except_block(f)
  File "exception_loosing_args.py", line 9, in try_except_block
    raise e
  File "exception_loosing_args.py", line 6, in try_except_block
    smthg.run()
  File "exception_loosing_args.py", line 22, in run
    re_call()
  File "exception_loosing_args.py", line 13, in re_call
    try_except_block(f2)
  File "exception_loosing_args.py", line 9, in try_except_block
    raise e
  File "exception_loosing_args.py", line 6, in try_except_block
    smthg.run()
  File "exception_loosing_args.py", line 20, in run
    raise KeyError
KeyError

我只打印一次＆＃34;错误＆＃34;。

在我的软件中，try_except_block函数被包装到一个对象中，我有想法设置这个对象的属性，该属性可以将异常语句标记为＆＃34;已经访问过＆＃34;或不。我不喜欢这个想法，因为这可能导致这种可能的副作用。还有另外一个吗？

示例我想到的（class属性由全局变量flag模拟）：

flag = False

def try_except_block(smthg):
   global flag
   try:
     smthg.run()
   except Exception as e:
     if not flag:
       print("WRONG")
       flag = True
     raise e

def re_call():
   f2 = Foo(True) # will throw an exception
   try_except_block(f2)

class Foo:
  def __init__(self, t):
    self.throw = t
  def run(self):
    if self.throw:
      raise KeyError
    else:
      re_call()

if __name__ == '__main__':
  f = Foo(False) # won't throw an exception
  try_except_block(f)

输出：

WRONG
Traceback (most recent call last):
  File "exception_loosing_args.py", line 28, in <module>
    try_except_block(f)
  File "exception_loosing_args.py", line 11, in try_except_block
    raise e
  File "exception_loosing_args.py", line 6, in try_except_block
    smthg.run()
  File "exception_loosing_args.py", line 24, in run
    re_call()
  File "exception_loosing_args.py", line 15, in re_call
    try_except_block(f2)
  File "exception_loosing_args.py", line 11, in try_except_block
    raise e
  File "exception_loosing_args.py", line 6, in try_except_block
    smthg.run()
  File "exception_loosing_args.py", line 22, in run
    raise KeyError
KeyError

Answer 1

我不完全理解为什么你的代码会做它的功能，但是可能适合的事情是将异常对象标记为已经看到，而不是设置全局标志。像这样：

def try_except_block(smthg):
    try:
        smthg.run()
    except Exception as e:
        if not hasattr(e, "teb_already_seen"):
            setattr(e, "teb_already_seen", True)
            print("WRONG")
        raise e

这样，try_except_block完成的额外处理只会发生一次每个异常对象，这可能是你想要的。

如何防止多次捕获异常

1 个答案: