python * operator, creating list with default value

Question

How do the following work?

>>> 3*[2]
>>> [2,2,2]
>>> [2]*3
>>> [2,2,2]

I understand that * is the positional expansion operator. Since [2] is a list with a single item, I don't see how 3*[2] expands to anything meaningful, I'd expect a SyntaxError, but that's not the case.

I'm having a hard time searching for an existing answer, all I find are references to *args and **kwargs for passing variadic parameter lists, which don't quite answer my question.

Answer 1

* is the multiplication operator. All Python sequences support multiplication. See the Sequence Types documentation:

s * n, n * s
n shallow copies of s concatenated

Note that the copies are shallow; any nested mutable types are not recursively copied too. This can lead to suprising results:

>>> nested = [[None]] * 5
>>> nested
[[None], [None], [None], [None], [None]]
>>> nested[0].append(42)
>>> nested
[[None, 42], [None, 42], [None, 42], [None, 42], [None, 42]]

There is just one nested [None] list, referenced 5 times, not 5 separate list objects.

The *args and **kw variable argument syntax only has meaning in a function definition or in a call (so callable_object(<arguments>)). It doesn't apply here at all. See What does ** (double star) and * (star) do for parameters? for more detail on that syntax.

Sequence types overload the * operator through the object.__mul__() and object.__rmul__() methods (when being the left or right operand in the expression), see Emulating container types for documentation on what hooks sequence types typically implement.