defining functions with optional arguments

Question

Hi I'm trying to understand how to implement optional arguments in a python function. For example, in the basic function below

def Ham(p,*q):
    if q:
        print p+q
    else:
        print p

Ham(2)


Ham(2,3)

I expect Ham(2) to return '2' which it does, however Ham(2,3) gives an error.

EDIT: Many thanks. Many of your answers were useful.

Have you read the relevant sections of [the Python tutorial](https://docs.python.org/2/tutorial/controlflow.html#more-on-defining-functions)? — BrenBarn, Jun 13 '15 at 22:28

score 2 · Accepted Answer · answered Jun 13 '15 at 22:38

2

In your particular example, I think you mean to do:

def Ham(p,q=None):
    if q:
        print p+q
    else:
        print p

That is, give q a default value of None and then only calculate p+q if a q is provided. Even simpler would be:

def Ham(p,q=0):
    print p+q

answered Jun 13 '15 at 22:38

xnx

This was a bad example. I was trying to make use of the optional argument. – piccolo Jun 13 '15 at 22:42
@Imran, don't dig in yourself deeper. `*q` doesn't do what you think it does, read the tutorial. – alexis Jun 13 '15 at 22:44

score 0 · Answer 2 · edited May 23 '17 at 12:07

0

Using *q you are specifyng a list of arguments not known until runtime and python expecting q to be a tuple, it means you can call Ham like below:

Ham(1, 2) # q = (2,)
Ham(1, 2, 3) # q = (2, 3)
Ham(1, 2, 3, 4) # q = (2, 3, 4)

Insidia of Ham function you have to treat q as a tuple, I mean, q[0] rather than q. You can have a look to this link to have a better idea.

edited May 23 '17 at 12:07

Community

answered Jun 13 '15 at 22:37

Israel Menis

2 Answers2