Python 3: Asterisk in print function

Question

Let's have a look:

print([object, ...], *, sep=' ', end='\n', file=sys.stdout)

http://docs.python.org/py3k/library/functions.html?highlight=print#print

How can we interpret that '*'? Usually an asterisk ('*') means numerous objects. But herein it is a mystery to me. Between two commas... I'm even afraid to think it may be a typo.

Answer 1

That's an error in the documentation, inserted by someone applying a new Python 3 feature to places where it shouldn't be used. It has since been fixed (see issue 15831).

The function signatures in the document used is given in a psuedo-formal-grammar form, but adding in the * marker only makes sense if you use actual python syntax. The [object, ...], * part of the signature should have been listed as *objects instead in that case.

The corrected version now reads:

print(*objects, sep=' ', end='\\n', file=sys.stdout, flush=False)

The online development version of the documentation has as of now not yet been updated, but the documentation source has been corrected; I'll see if we can request a regeneration of the docs.

To be clear: the * syntax is valid in Python 3 and means that the following arguments can only ever be used as keyword arguments, not positional arguments. This does however not apply to the print() function, as all positional arguments are to be printed anyway and could never be mistaken for the keyword arguments.

Answer 2

It means that the following arguments are keyword-only i.e., you can't supply them as positional arguments, you must use their names e.g.:

>>> def f(*, a): pass
... 
>>> f(1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: f() takes exactly 0 positional arguments (1 given)
>>> f(a=1)
>>> # ok

Another example:

>>> def g(*a, b): pass
... 
>>> g(1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: g() needs keyword-only argument b
>>> g(1, 2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: g() needs keyword-only argument b
>>> g(1, b=2)
>>> # ok
>>> g(1, 2, b=3)
>>> # ok