Adding information to an exception?

Question

I want to achieve something like this:

def foo():
   try:
       raise IOError(\'Stuff \')
   except:
       raise

def bar(arg1):
    try:
       foo()
             


        

Answer 1

Unlike previous answers, this works in the face of exceptions with really bad __str__. It does modify the type however, in order to factor out unhelpful __str__ implementations.

I'd still like to find an additional improvement that doesn't modify the type.

from contextlib import contextmanager
@contextmanager
def helpful_info():
    try:
        yield
    except Exception as e:
        class CloneException(Exception): pass
        CloneException.__name__ = type(e).__name__
        CloneException.__module___ = type(e).__module__
        helpful_message = '%s\n\nhelpful info!' % e
        import sys
        raise CloneException, helpful_message, sys.exc_traceback


class BadException(Exception):
    def __str__(self):
        return 'wat.'

with helpful_info():
    raise BadException('fooooo')

The original traceback and type (name) are preserved.

Traceback (most recent call last):
  File "re_raise.py", line 20, in 
    raise BadException('fooooo')
  File "/usr/lib64/python2.6/contextlib.py", line 34, in __exit__
    self.gen.throw(type, value, traceback)
  File "re_raise.py", line 5, in helpful_info
    yield
  File "re_raise.py", line 20, in 
    raise BadException('fooooo')
__main__.BadException: wat.

helpful info!