unpacking function argument
function, python
Solution
You are looking for the `*args` argument syntax:
>>> def foo(bar, baz, spam):
... print bar, baz, spam
...
>>> arguments = [1, 2, 3]
>>> foo(*arguments)
1, 2, 3
When passing arguments to a callable, any expression preceded by a `*` asterix, is interpreted as a sequence of positional arguments, and expanded to be passed on as separate arguments to the called object (function, method, etc.).
For your example that would be:
func1(*func2(...))
There is a keyword equivalent using `**` double asterixes (takes a mapping), and you can use the same syntax in function signatures too.
See the documentation on call expressions, and for the function signature mirror syntax, the documentation on function definitions.
Problem
if a function takes n number of arguments, and there is another function that returns a sequence with n number of items(or I have a sequence with n number of items), is there a way to 'map' these two functions(or make the first function take a sequence of n number of items as input and return result) I don't want (and maybe can't in some occasions) to edit the function myself and change its arguments parameters and return value types. i.e) ``` def func1(x, y, z): return x+y+z def func2(w): return [i for i in range(w,w+3)] ``` cant `func1(func2( ... ))` in this case.