Python - value unpacking order in method parameters

Question

def fun(a, b, c, d):
    print('a:', a, 'b:', b, 'c:', c, 'd:', d)

why this one works

fun(3, 7, d=10, *(23,))

and prints out:

a: 3 b: 7 c: 23 d: 10

while this

fun(3, 7, c=10, *(23,))

does not

Traceback (most recent call last):
  File "/home/lookash/PycharmProjects/PythonLearning/learning.py", line 10, in <module>
    fun(3, 7, c=10, *(23,))
TypeError: fun() got multiple values for argument 'c'

Answer 1

With *(23,) , you are unpacking the values in the tuple (23,) as positional arguments, following the positional arguments that are already defined, namely 3 for a and 7 for b , so 23 would be assigned to parameter c , which is why fun(3, 7, d=10, *(23,)) works, but in fun(3, 7, c=10, *(23,)) you are also assigning to value 10 to c as a keyword argument, so it is considered a conflict as c cannot be assigned with both 23 and 10 .

Note that while legal, it is discouraged by some to unpack iterable arguments after keyword arguments, as discussed here , although the syntax is ultimately ruled to stay.