Passing tuple to a function

Question

In the following code:

def fun(args, **kwargs):
    print(len(args))

thread = threading.Thread(
    target=fun, args=('Cats', 'Dogs', 'Tigers'), kwargs={'sep': ' & '})

thread.start()

I am trying to pass a tuple to the parameter args but it gives the following error:

TypeError: fun() takes 1 positional argument but 3 were given

But when I use args=(('Cats', 'Dogs', 'Tigers'),) it works. I understand that it is considered as a tuple but why ('Cats', 'Dogs', 'Tigers') is not being considered as a tuple?

Answer 1

In the first case you are passing a tuple as Thread's args parameter. In the second case you are passing a tuple containing a tuple as its first entry.

len(
  ('Cats', 'Dogs', 'Tigers')
) # 3

len(
  (('Cats', 'Dogs', 'Tigers'))
) # 1

The args parameter of threading.Thread is splatted into target's call (so each argument in the iterable passed to args is unpacked into a separate argument for target):

threading.Thread(target=fun, args=('Cats', 'Dogs', 'Tigers'))
# is equivalent to (but in another thread)
fun(*('Cats', 'Dogs', 'Tigers'))
# which is equivalent to
fun('Cats', 'Dogs', 'Tigers')

compared to:

threading.Thread(target=fun, args=(('Cats', 'Dogs', 'Tigers'),))
# is equivalent to (but in another thread)
fun(*(('Cats', 'Dogs', 'Tigers'),))
# which is equivalent to
fun(('Cats', 'Dogs', 'Tigers'))