将带有args的方法传递给带有kwargs的另一个函数

Question

I have two functions that differ by one argument that I would like to pass in accordingly based on a given case...

def func1(a, b, c):
    print a, b, c

def func2(a, b, c, d=False)
    print a, b, c, d

def run(func, **kwargs):
     if b is None:
         b = 999
     func(**kwargs)

run(func1, a=1, b=None, c=3)

I am unable to get this to work as it complains that b is being referenced before assignment.

Answer 1

def func1(a, b, c):
    print a, b, c

def func2(a, b, c, d=False):
    print a, b, c, d

def run(func, **kwargs):
    if 'b' in kwargs:
        if kwargs['b'] is None:
            kwargs['b'] = 999
    func(**kwargs)

run(func1, a=1, b=None, c=3)

Answer 2

Access b by:

if kwargs['b'] is None:

So, The run function would be:

def run(func, **kwargs):
     kwargs['b'] = kwargs.get('b', 999)
     func(**kwargs)