Number of actual function arguments

Question

Is there a way to get the number of actual parameters passed to a function

def foo(a, optional=42):
    if ????
        print "1 arg"
    else:
        print "2 args"

foo(11)      # should print 1 arg
foo(22, 42)  # should print 2 args

without changing its signature to accept *args?

Answer 1

You could change the default value to a sentinel:

_sentinel = object()

def foo(a, optional=_sentinel):
    if optional is _sentinel:
        optional = 42
        print "1 arg"
    else:
        print "2 args"

or by accessing it directly in the func_defaults tuple:

def foo(a, optional=object()):
    if optional is foo.func_defaults[0]:
        optional = 42
        print "1 arg"
    else:
        print "2 args"

but don't actually use that; that'll just serve to confuse those not familiar with standard function object attributes.

Yes, the _sentinel object is introspectable and can be obtained still by a determined developer, but then again that same developer could just monkeypatch your function as well. :-)

Answer 2

You can use a decorator to do that. From the Preserving signatures of decorated functions question we know how to do this correctly.

import decorator


@decorator.decorator
def count_args(f):
    def new(*args, **kwargs):
        kwargs['nargs'] = len(args)+len(kwargs)
        return f(*args, **kwargs)
    return new

@count_args
def foo(a, optional=42, nargs=1):
    print nargs

foo(1) # 1
foo(1, 4) # 2
foo(1, optional=4) # 2

Update:

I have just added keyword argument for the number of args, that we passed to the function. As you can see, by default it is 1, which is True. This looks like a hack, but it works.