如何在不使用 splat 运算符的情况下解压缩具有不同数量元素的元组？

Question

It is a part of my university project, and I am trying to make a function that returns multiple variables (unpacked tuple) to call another function with a proper number of arguments.

value = function_list[i](*func())

This was easy just using *(splat) operator until one of our instructors told us not to use this operator. Then, I tried to unpack tuple by assigning it into variables manually.

a, b = func()
function_list[i](a, b) #Causes error when a function requires more/less than 2 arguments.

Function_list is a list of multiple function definitions that may have a different number of parameters. So this is a problem in some cases. Those functions are not made by myself therefore I can't change those parameters to be optional.

Is there any finest way to replace * operator to get my all functions working properly?

Also, I am not allowed to use these keywords:

as, assert, asynch, await, break, class, continue, except, finally, global, is, lambda, nonlocal, raise, try, with, yield. 

Answer 1

I think it is a weird requirement not to be able to use * , but here's what I've come up with:

import inspect, functools

def get_min_max_arg_count(func):
    params = inspect.signature(func).parameters
    defaults = [p.default for p in params.values()]
    return defaults.count(inspect._empty), len(params)

def call_with_params(func, params):
    for param in params:
        func = functools.partial(func, param)
    return func()

def call_matching_func(func, function_list):
    '''Call `func` and the first function in `function_list` that
       has a signature matching the tuple returned by `func`.'''
    params = func()
    n_params = len(params)
    for fn in function_list:
        n_req, n_max = get_min_max_arg_count(fn)
        if n_req <= n_params <= n_max:
            return call_with_params(fn, params)

Note that call_matching_func assumes that any func you'll pass in returns an iterable.

Note also that I've decided that keyword arguments do not exist.