Question

I want to know how to call a function, passing a parameter that it might not be expecting.

I faced this problem a few days ago, and found a way around it. But today, I decided I’d see if what I wanted to do was possible. Unfortunately, I don’t remember the context which I used it in. So here is a stupid example in which there are plenty of better ways to do this, but just ignore that:

def test(func, arg1, arg2):
    return func(arg1, arg2, flag1=True, flag2=False) #Only pass the flags if the function accepts them.

def func1(a, b, flag1, flag2):
    ret = a
    if flag1:
        ret *= b
    if flag2:
        ret += b
    return ret

def func2(a, b):
    return a*b

print test(func1, 5, 6) #prints 30

The alternative I came up with looked like this:

def test(func, arg1, arg2):
    numArgs = len(inspect.getargspec(func).args)
    if numArgs >= 4:
        return func(arg1, arg2, True, False)
    elif numArgs == 3:
        return func(arg1, arg2, True)
    else:
        return func(arg1, arg2)

print test(func2, 5, 6) #prints 30

or a try..except.. block

But there’s got to be a better way of doing this without altering func1 and func2, right?

(Edit): Making use of the solution provided by Max S, I’m thinking this is the best approach:

def callWithOptionalArgs(func, *args):
    argSpec = inspect.getargspec(func)
    lenArgSpec = len(argSpec.args or ())
    argsToPass = args[:lenArgSpec] #too many args
    defaults = argSpec.defaults or ()
    lenDefaults = len(defaults)
    argsToPass += (None, )*(lenArgSpec-len(argsToPass)-lenDefaults) #too few args
    argsToPass += defaults[len(argsToPass)+len(defaults)-lenArgSpec:] #default args

    return func(*argsToPass)
print callWithOptionalArgs(func1, 5, 6, True) #prints 30
print callWithOptionalArgs(func2, 5, 6, True) #prints 30

Answer 1

Inspecting the function is the only way to explicitly differentiate between functions with different numbers of arguments without altering or decorating the originals. The only change I would do to your wrapper is to generalize it for any number of arguments:

def padArgsWithTrue(func, *args):
    passed_args = list(args)
    num_args = len(inspect.getargspec(func).args)
    passed_args += [True] * (num_args - len(args))
    return func(*passed_args)

print padArgsWithTrue(lambda x,y,z,w: (x*y, z, w), 5, 6)

EDIT: Note that this does not accommodate functions with variable number of args or keyword args. You’ll have to decide on a policy to deal with those before a complete solution could be written.