如何在没有堆栈跟踪的情况下创建 Python 异常？

Question

I would like to create a Python exception that does not print the stack trace (for simple user errors). I have created a function below that wraps a custom class, but I would prefer to bake it into the custom exception class directly, would that be possible somehow?

class UserError(Exception):
    pass

def throw_user_error(s_msg):
    sys.tracebacklimit = 0
    raise UserError(s_msg)
    sys.tracebacklimit = 1000

Answer 1

To achieve that, you should create custom handler, not the exception. And this handler even already exists in the standard library: For example:

class CustomException(Exception):
    pass

from contextlib import suppress

def main():
    with suppress(CustomException):
        # throw this error without consequences
        raise CustomException() 
        print('This won\'t be reached')   
    print('Exiting normally anyway')

main()    

---

Exiting normally anyway

And one more effective way to hide traceback is to replace it:

In [30]: def fn(n=0):
    ...:     if n <= 0:
    ...:         raise Exception
    ...:     return fn(n-1)
    ...:
    ...: try:
    ...:     fn(10)
    ...: except Exception as e:
    ...:     raise e.with_traceback(None)
---------------------------------------------------------------------------
Exception                                 Traceback (most recent call last)
<ipython-input-30-bc3c4e86e3cc> in <module>
      7     fn(10)
      8 except Exception as e:
----> 9     raise e.with_traceback(None)

Exception:

Without this line (believe me or not) if we propagate the exception further, we'll see the long-long stack of recursive calls.