Can a variable number of arguments be passed to a function?

Question

In a similar way to using varargs in C or C++:

fn(a, b)
fn(a, b, c, d, ...)

Answer 1

Yes. You can use *args as a non-keyword argument. You will then be able to pass any number of arguments.

def manyArgs(*arg):
  print "I was called with", len(arg), "arguments:", arg

>>> manyArgs(1)
I was called with 1 arguments: (1,)
>>> manyArgs(1, 2, 3)
I was called with 3 arguments: (1, 2, 3)

As you can see, Python will unpack the arguments as a single tuple with all the arguments.

For keyword arguments you need to accept those as a separate actual argument, as shown in Skurmedel's answer.

Answer 2

Adding to unwinds post:

You can send multiple key-value args too.

def myfunc(**kwargs):
    # kwargs is a dictionary.
    for k,v in kwargs.iteritems():
         print "%s = %s" % (k, v)

myfunc(abc=123, efh=456)
# abc = 123
# efh = 456

And you can mix the two:

def myfunc2(*args, **kwargs):
   for a in args:
       print a
   for k,v in kwargs.iteritems():
       print "%s = %s" % (k, v)

myfunc2(1, 2, 3, banan=123)
# 1
# 2
# 3
# banan = 123

They must be both declared and called in that order, that is the function signature needs to be *args, **kwargs, and called in that order.

Answer 3

If I may, Skurmedel's code is for python 2; to adapt it to python 3, change iteritems to items and add parenthesis to print. That could prevent beginners like me to bump into: AttributeError: 'dict' object has no attribute 'iteritems' and search elsewhere (e.g. Error “ 'dict' object has no attribute 'iteritems' ” when trying to use NetworkX's write_shp()) why this is happening.

def myfunc(**kwargs):
for k,v in kwargs.items():
   print("%s = %s" % (k, v))

myfunc(abc=123, efh=456)
# abc = 123
# efh = 456

and:

def myfunc2(*args, **kwargs):
   for a in args:
       print(a)
   for k,v in kwargs.items():
       print("%s = %s" % (k, v))

myfunc2(1, 2, 3, banan=123)
# 1
# 2
# 3
# banan = 123