In python, I can define a function as follows:
def func(kw1=None,kw2=None,**kwargs):
...
In this case, i can call func as:
func(kw1=3,kw2=4,who_knows_if_this_will_be_used=7,more_kwargs=Ellipsis)
I can also define a function as:
def func(arg1,arg2,*args):
...
which can be called as
func(3,4,additional,arguments,go,here,Ellipsis)
Finally, I can combine the two forms
def func(arg1,arg2,*args,**kwargs):
...
But, what does not work is calling:
func(arg1,arg2,*args,kw1=None,kw2=None,**kwargs): #SYNTAX ERROR (in python 2 only, apparently this works in python 3)
...
My original thought was that this was probably because a function
def func(arg1,arg2,*args,kw1=None):
...
can be called as
func(1,2,3) #kw1 will be assigned 3
So this would introduce some ambiguity as to whether 3 should be packed into args or kwargs. However, with python 3, there is the ability to specify keyword only arguments:
def func(a,b,*,kw=None): #can be called as func(1,2), func(1,2,kw=3), but NOT func(1,2,3)
...
With this, it seems that there is no syntactic ambiguity with:
def func(a,b,*args,*,kw1=None,**kwargs):
...
However, this still brings up a syntax error (tested with Python3.2). Is there a reason for this that I am missing? And, is there a way to get the behavior I described above (Having *args with default arguments) -- I know I can simulate that behavior by manipulating the kwargs dictionary inside the function.
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