错误发生时运行功能，无需尝试

Question

I am trying to run a function whenever an error occurs. I don't want to use try or except because my code is very large and there are so much chances of an error to occur , so I can't use try .. everywhere. This is what I am expecting:

>>> if ValueError:  #don't works , just assuming.
    print("Hey! You are done.")

>>> int("abc")
Hey! You are done.
>>> int("1b")
Hey! You are done.
>>> 

Is there any way to do this?

Answer 1

If your code is a whole block, I recommend splitting it into functions. From here, you can wrap each function in a decorator that takes an error and a function to be run on error:

def handle_error(error, error_func):
    def decorator(func):
        def wrapper(*args, **kwargs):
            r = None
            try:
                r = func(*args, **kwargs)
            except error as e:
                r = error_func()
            finally:
                return r
        return wrapper
    return decorator

Then use on functions like so:

def bad_value():
    print('bad value given!')

@handle_error(ValueError, bad_value)
def multiply(a, b):
    return a * b

Of course, you can be more 'broad', and catch all exceptions...

@handle_error(Exception, error_func)
def func(): 
    # ...

Answer 2

A ValueError is triggered by a particular piece of code, and will raise itself immediately after the offending statement. int("abc") will raise a ValueError on its own, and the program execution will stop, before it reaches any if ValueError statement.

You need a try / except block to allow python to catch the error and continue executing. I can't see any way to achieve what you want without one.