[SOLVED] How do I create a Python function with optional arguments?

Issue

I have a Python function which takes several arguments. Some of these arguments could be omitted in some scenarios.

def some_function (self, a, b, c, d = None, e = None, f = None, g = None, h = None):
    #code

The arguments d through h are strings which each have different meanings. It is important that I can choose which optional parameters to pass in any combination. For example, (a, b, C, d, e), or (a, b, C, g, h), or (a, b, C, d, e, f, or all of them (these are my choices).

It would be great if I could overload the function – but I read that Python does not support overloading. I tried to insert some of the required int arguments in the list – and got an argument mismatch error.

Right now I am sending empty strings in place of the first few missing arguments as placeholders. I would like to be able to call a function just using actual values.

Is there any way to do this? Could I pass a list instead of the argument list?

Right now the prototype using ctypes looks something like:

_fdll.some_function.argtypes = [c_void_p, c_char_p, c_int, c_char_p, c_char_p, c_char_p, c_char_p, c_char_p]

Solution

Try calling it like: obj.some_function( '1', 2, '3', g="foo", h="bar" ). After the required positional arguments, you can specify specific optional arguments by name.

Answered By – Russell Borogove

Answer Checked By – Cary Denson (BugsFixing Admin)

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