I know that I can use *args to define a function with an arbitrary number of arguments. My question is a little bit different: What if I want the number of arguments to be controlled by a variable? For example, the number of arguments should be 2n, where the value of n is calculated earlier in the code?
1 Answer
You can get the number of arguments when using *args with len (because args is a tuple) and act based on that (included some test cases):
number = 2
def func(*args):
if len(args) != number * 2:
raise TypeError(f'Expected {number * 2} arguments, got {len(args)}')
# do some other stuff
# else clause not needed
# testing
test_cases = [
(1, 2, 3),
(1, 2, 3, 4),
(1, 2, 3, 4, 5)
]
for arguments in test_cases:
print(f'Calling function with {len(arguments)} arguments')
try:
func(*arguments)
except TypeError as e:
print(f'Raised an exception: {e}')
else:
print('Executed without exceptions')
print()
# output:
# Calling function with 3 arguments
# Raised an exception: Expected 4 arguments, got 3
#
# Calling function with 4 arguments
# Executed without exceptions
#
# Calling function with 5 arguments
# Raised an exception: Expected 4 arguments, got 5
*args, thenargsis a list, so you can uselen(args)(inside the function obvs) and raise aTypeErrorif the length doesn't match some predetermined valueassert len(args) == 2*nto the first line of the function.assertmethod mentioned by @KotaMori