function with variable arguments

[Peter Dembinski]

On Fri, 13 May 2005 11:52:34 +0200, Xah Lee <xah at xahlee.org> wrote:

> i wanted to define a function where the number of argument matters.
> Example:
>
> def Range(n):
>     return range(n+1)
>
> def Range(n,m):
>     return range(n,m+1)
>
> def Range(n,m,step):
>     return range(n,m+1,step)
>
> this obvious doesn't work. The default argument like
> Range(n=1,m,step=1) obviously isn't a solution.
>
> can this be done in Python?
>
> or, must the args be changed to a list?

It can be written this way:

   def Range_3args(n, m, step):
       return range(n, m + 1, step)

   def Range_2args(n, m):
       return range(n, m + 1)

   def Range(n, m = None, step = None):
       if (m is None) and (step is None):
           return range(n + 1)

       if (not (m is None)) and (step is None):
           return Range_2args(n, m)

       if (not (m is None)) and (not (step is None)):
           return Return_3args(n, m, step)

       return []


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