Question: I am searching for a way to change the printable output of an Exception to a silly message in order to learn more about python internals (and mess with a friend ;), so far without success. Consider the following code try: x # is not defined except NameError as exc: print(exc)  The code shall output name 'x' is not defined I would like the change that output to the name 'x' you suggested is not yet defined, my lord. Improve your coding skills. So far, I understood that you can't change __builtins__ because they're "baked in" as C code, unless:  You use forbiddenfruit.curse method which adds / changes properties of any object You manually override the dictionnaries of an object  I've tried both solutions, but without success: forbiddenfruit solution: from forbiddenfruit import curse curse(BaseException, 'repr', lambda self: print("Test message for repr")) curse(BaseException, 'str', lambda self: print("Test message for str")) try: x except NameError as exc: print(exc.str()) # Works, shows test message print(exc.repr()) # Works, shows test message print(repr(exc)) # Does not work, shows real message print(str(exc)) # Does not work, shows real message print(exc) # Does not work, shows real message  Dictionnary overriding solution: import gc underlying_dict = gc.get_referents(BaseException.__dict__)[0] underlying_dict["__repr__"] = lambda self: print("test message for repr") underlying_dict["__str__"] = lambda self: print("test message for str") underlying_dict["args"] = 'I am an argument list' try: x except NameError as exc: print(exc.__str__()) # Works, shows test message print(exc.__repr__()) # Works, shows test message print(repr(exc)) # Does not work, shows real message print(str(exc)) # Does not work, shows real message print(exc) # Does not work, shows real message  AFAIK, using print(exc) should rely on either __repr__ or __str__, but it seems like the print function uses something else, which I cannot find even when reading all properties of BaseException via print(dir(BaseException)). Could anyone give me an insight of what print uses in this case please ? [EDIT] To add a bit more context: The problem I'm trying to solve began as a joke to mess with a programmer friend, but now became a challenge for me to understand more of python's internals. There's no real business problem I'm trying to solve, I just want to get deeper understanding of things in Python. I'm quite puzzled that print(exc) won't make use of BaseException.__repr__ or __str__ actually. [/EDIT]

Answer: I'll just explain the behaviour you described:  exc.__repr__()  This will just call your lambda function and return the expected string. Btw you should return the string, not print it in your lambda functions.  print(repr(exc))  Now, this is going a different route in CPython and you can see this in a GDB session, it's something like this: Python/bltinmodule.c:builtin_repr will call Objects/object.c:PyObject_Repr - this function gets the PyObject *v as the only parameter that it will use to get and call a function that implements the built-in function repr(), BaseException_repr in this case. This function will format the error message based on a value from args structure field: (gdb) p ((PyBaseExceptionObject *) self)->args $188 = ("name 'x' is not defined",)  The args value is set in Python/ceval.c:format_exc_check_arg based on a NAME_ERROR_MSG macro set in the same file. Update: Sun 8 Nov 20:19:26 UTC 2020 test.py: import sys import dis def main(): try: x except NameError as exc: tb = sys.exc_info()[2] frame, i = tb.tb_frame, tb.tb_lasti code = frame.f_code arg = code.co_code[i + 1] name = code.co_names[arg] print(name) if __name__ == '__main__': main()  Test: # python test.py x  Note: I would also recommend to watch this video from PyCon 2016.