How to quit Python function, throwing error statement without quitting Python interpreter

Question

I'm new to Python and struggling with handling self-defined errors. When my code spots the error, I want it to throw an error in red font and take me back to the Python terminal without killing Python.

I came across sys.exit() looking for an answer, but it quits Python completely. Do you know of an alternative that throws back an error in red font and takes me back to the terminal?

This is what I have so far.

import sys def do_something(parameter):     if parameter > 100:         # quit the function and any function(s) that may have called it         sys.exit('Your parameter should not be greater than 100!')     else:         # otherwise, carry on with the rest of the code 

Please let me know if I'm not clear and I'll be happy to provide more details. Thank you all in advance!

Answer 1

You have two options (at least).

Using a return statement:

def do_something(parameter):     if parameter > 100:         # display error message if necessary         return  # 'exit' function and return to caller     # rest of the code 

You can also return soemthing passing the something value back to the caller. This can be used to provide a status code for instance (e.g. 0: success, 1: error).

Or a better approach is to raise an exception:

def do_something(parameter):     if parameter > 100:         raise ValueError('Parameter should...')     # rest of the code  try:     do_something(101) except ValueError, e:     # display error message if necessary e.g. print str(e) 

See exceptions in the Python manual.

There are built-in exception classes (like ValueError above). You can also define your own as follows:

class ParameterError(Exception):     pass 

You can also add additional code to your custom exception classes to process parameters, display custom error messages, etc...

The built-in exceptions are listed here.