Trying to understand optional, list and named arguments in Python -

i'm beginner @ python, trying understand function arguments , types , orders.

i tried experiment little different kinds of argument , here's experiment

def main():     test_a(2, 33, 44)     test_b(2, 33)     test_c(2)     ## test_d(2,,44) **produces invalid syntax**      test_e(2,33,44,55,66)     test_f(2, 44,55,66, y = 44)      test_g(2, 33, 44,55,66, rofa = 777, nard = 888)     ##test_h(2, 33, foo = 777, boo = 888, 44,55,66)  **invalid syntax in function definition     ##test_l(2, 44,55,66 ,  foo= 777, boo = 888, y  = 900) **invalid syntax in function definition     test_m(2, 44,55,66 , y = 900, foo=77777 , boo = 88888)  ############################################################# ## no optional arguments def test_a(x,y,z):     print("test_a : x = {}, y = {}, z = {} ".format(x ,y ,z))  ## 1 optional argument @ end def test_b(x, y, z = 22):     print("test_b : x = {}, y = {}, z = {} ".format(x ,y ,z))  ## 2 optional arguments @ end def test_c(x, y = 11, z = 22):     print("test_c : x = {}, y = {}, z = {} ".format(x ,y ,z))  ## 1 optional argument @ middle ## produces non-default argument follows default argument #### **** default arguments must come @ end **** #### ## def test_d(x, y = 11, z): ##    print("test_d : x = {}, y = {}, z = {} ".format(x ,y ,z))  #################################################################  ## no optional argument + 1 list argument def test_e(x, y, *args):     print("test_e : x = {}, y = {} ||".format(x, y), end= " ")     in args :         print(i)  ## 1 optional argument + 1 list argument def test_f(x, *args , y = 5):     print("test_f : x = {}, y = {} ||".format(x, y), end= " ")     in args :         print(i)  ################################################################  ## no optional argument, 1 list, 1 keyword arguments def test_g(x,y,*args, **kwargs):     print(x, y)     in args:         print(i)      i, v in kwargs.items():         print(i, v)  ## **kwargs befor *args produces syntax error !!! ##def test_h(x,y, **kwargs, *args): ##    print(x, y) ##    in args: ##        print(i) ## ##    i, v in kwargs.items(): ##        print(i, v)  ## **kwargs befor optional argument produces syntax error !!! ##def test_l(x,*args,**kwargs, y = 5): ##    print(x, y) ##    in args: ##        print(i) ## ##    i, v in kwargs.items(): ##        print(i, v) ##      ## 1 optiona, list , keyword arguments def test_m(x,*args,y = 5, **kwargs):     print(x, y)     in args:         print(i)      i, v in kwargs.items():         print(i, v)   if __name__ == "__main__":     main() 

i understood of things after experiment. there's 1 issue can't inside head.

in function definition in test_h , test_m **kwargs defined before optional argument , list argument, when run program, if didn't use function, defined it.. produces syntax error .. i'd grateful know why happening ?

thanks.

keyword arguments have final arguments passed function. read on here. parameters **kwargs must last parameter in function signature (as detailed on page in docs). documentation:

when final formal parameter of form **name present, receives dictionary (see mapping types — dict) containing keyword arguments except corresponding formal parameter.

the reason code raises syntaxerror if don't use function if violate rule, python can't complete function definition since signature you're trying give illegal.