Understanding *x ,= lst

Question

I\'m going through some old code trying to understand what it does, and I came across this odd statement:

*x ,= p

p is a list in t

Answer 1

You should always throw these to dis and see what it throws back at you; you'll see how *x, = p is actually different from x = p:

dis('*x, = p')
  1           0 LOAD_NAME                0 (p)
              2 UNPACK_EX                0
              4 STORE_NAME               1 (x)

While, the simple assignment statement:

dis('x = p')
  1           0 LOAD_NAME                0 (p)
              2 STORE_NAME               1 (x)

^{(Stripping off unrelated None returns)}

As you can see UNPACK_EX is the different op-code between these; it's documented as:

Implements assignment with a starred target: Unpacks an iterable in TOS (top of stack) into individual values, where the total number of values can be smaller than the number of items in the iterable: one of the new values will be a list of all leftover items.

Which is why, as Eugene noted, you get a new object that's referred to by the name x and not a reference to an already existing object (as is the case with x = p).

---

*x, does seem very odd (the extra comma there and all) but it is required here. The left hand side must either be a tuple or a list and, due to the quirkiness of creating a single element tuple in Python, you need to use a trailing ,:

i = 1, # one element tuple

If you like confusing people, you can always use the list version of this:

[*x] = p

which does exactly the same thing but doesn't have that extra comma hanging around there.