function call - default value & collecting arguments

[OKB (not okblacke)]

Primoz Skale wrote:

> OK, everything allright till so fair. But! :) Now define third
> function as: 
> 
> def f(*a):
>    print a[0]
> 
> In this case, function only prints first argument in the tuple:
> 
>>>f(1,2,3)
> 1
>>>f(3)
> 3
>>>f()    #no arguments passed
> Traceback (most recent call last):
>   File "<pyshell#425>", line 1, in <module>
>     f() #no arguments passed
>   File "<pyshell#422>", line 2, in f
>     print a[0]
> IndexError: tuple index out of range
> 
> Then I tried to define the function as:
> 
> def f(*a=(0,)):
>   print a[0]  #this should come next, but we get error msg instead,
>   saying 
> 
>           SyntaxError: invalid syntax
> 
> but it does not work this way. Now my 'newbie' question: Why not?
> :) I wanted to write function in this way, because then we can call
> function without any arguments, and it would still print 0 (in this
> case). 
> 
> What I wanted was something like this:
> 
> def f(*a=(0,)):
>   print a[0]
> 
>>>f(1,2)
> 1
>>>f()  #no args passed 0

    	When you use the *a syntax, you're saying that you want a to 
contain a tuple of all the arguments.  When you try *a=(0,), you seem to 
be saying that you want a to hold all the arguments, or, if there are 
none, you want to pretend that it was called with one argument, namely 
0.

    	Well, if you want to ensure that there is at least one argument, 
and print that first one, and make it zero if it's not supplied, why are 
you using the *a syntax?  You're clearly treating that first argument 
distinctly, since you want to apply a default value to it but not to any 
others.  Just do this:

def f(a, *b):
    	print a

-- 
--OKB (not okblacke)
Brendan Barnwell
"Do not follow where the path may lead.  Go, instead, where there is
no path, and leave a trail."
	--author unknown