Generator as function argument

Question

Can anyone explain why passing a generator as the only positional argument to a function seems to have special rules?

If we have:

def f(*args):
    print "Success!"
    print args

1. This works, as expected.

   >>> f(1, *[2])
   Success!
   (1, 2)
   

2. This does not work, as expected.

   >>> f(*[2], 1)
     File "<stdin>", line 1
   SyntaxError: only named arguments may follow *expression
   

3. This works, as expected

   >>> f(1 for x in [1], *[2])
   Success! 
   (generator object <genexpr> at 0x7effe06bdcd0>, 2)
   

4. This works, but I don't understand why. Shouldn't it fail in the same way as 2)

   >>> f(*[2], 1 for x in [1])
   Success!
   (generator object <genexpr> at 0x7effe06bdcd0>, 2)
   

Solution

Both 3. and 4. should be syntax errors on all Python versions. However you've found a bug that affects Python versions 2.5 - 3.4, and which was subsequently posted to the Python issue tracker. Because of the bug, an unparenthesized generator expression was accepted as an argument to a function if it was accompanied only by *args and/or **kwargs. While Python 2.6+ allowed both cases 3. and 4., Python 2.5 allowed only case 3. - yet both of them were against the documented grammar:

call    ::=     primary "(" [argument_list [","]
                            | expression genexpr_for] ")"

i.e. the documentation says a function call comprises of primary (the expression that evaluates to a callable), followed by, in parentheses, either an argument list or just an unparenthesized generator expression; and within the argument list, all generator expressions must be in parentheses.

---

This bug (though it seems it had not been known), had been fixed in Python 3.5 prereleases. In Python 3.5 parentheses are always required around a generator expression, unless it is the only argument to the function:

Python 3.5.0a4+ (default:a3f2b171b765, May 19 2015, 16:14:41) 
[GCC 4.9.2] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> f(1 for i in [42], *a)
  File "<stdin>", line 1
SyntaxError: Generator expression must be parenthesized if not sole argument

This is now documented in the What's New in Python 3.5, thanks to DeTeReR spotting this bug.

---

Analysis of the bug

There was a change made to Python 2.6 which allowed the use of keyword arguments after *args:

It’s also become legal to provide keyword arguments after a *args argument to a function call.

>>> def f(*args, **kw):
...     print args, kw
...
>>> f(1,2,3, *(4,5,6), keyword=13)
(1, 2, 3, 4, 5, 6) {'keyword': 13}

Previously this would have been a syntax error. (Contributed by Amaury Forgeot d’Arc; issue 3473.)

---

However, the Python 2.6 grammar does not make any distinction between keyword arguments, positional arguments, or bare generator expressions - they are all of type argument to the parser.

As per Python rules, a generator expression must be parenthesized if it is not the sole argument to the function. This is validated in the Python/ast.c:

for (i = 0; i < NCH(n); i++) {
    node *ch = CHILD(n, i);
    if (TYPE(ch) == argument) {
        if (NCH(ch) == 1)
            nargs++;
        else if (TYPE(CHILD(ch, 1)) == gen_for)
            ngens++;
        else
            nkeywords++;
    }
}
if (ngens > 1 || (ngens && (nargs || nkeywords))) {
    ast_error(n, "Generator expression must be parenthesized "
              "if not sole argument");
    return NULL;
}

However this function does not consider the *args at all - it specifically only looks for ordinary positional arguments and keyword arguments.

Further down in the same function, there is an error message generated for non-keyword arg after keyword arg:

if (TYPE(ch) == argument) {
    expr_ty e;
    if (NCH(ch) == 1) {
        if (nkeywords) {
            ast_error(CHILD(ch, 0),
                      "non-keyword arg after keyword arg");
            return NULL;
        }
        ...

But this again applies to arguments that are not unparenthesized generator expressions as evidenced by the else if statement:

else if (TYPE(CHILD(ch, 1)) == gen_for) {
    e = ast_for_genexp(c, ch);
    if (!e)
        return NULL;
    asdl_seq_SET(args, nargs++, e);
}

Thus an unparenthesized generator expression was allowed to slip pass.

---

Now in Python 3.5 one can use the *args anywhere in a function call, so the Grammar was changed to accommodate for this:

arglist: argument (',' argument)*  [',']

and

argument: ( test [comp_for] |
            test '=' test |
            '**' test |
            '*' test )

and the for loop was changed to

for (i = 0; i < NCH(n); i++) {
    node *ch = CHILD(n, i);
    if (TYPE(ch) == argument) {
        if (NCH(ch) == 1)
            nargs++;
        else if (TYPE(CHILD(ch, 1)) == comp_for)
            ngens++;
        else if (TYPE(CHILD(ch, 0)) == STAR)
            nargs++;
        else
            /* TYPE(CHILD(ch, 0)) == DOUBLESTAR or keyword argument */
            nkeywords++;
    }
}

Thus fixing the bug.

However the inadvertent change is that the valid looking constructions

func(i for i in [42], *args)

and

func(i for i in [42], **kwargs)

where an unparenthesized generator precedes *args or **kwargs now stopped working.

---

To locate this bug, I tried various Python versions. In 2.5 you'd get SyntaxError:

Python 2.5.5 (r255:77872, Nov 28 2010, 16:43:48) 
[GCC 4.4.5] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> f(*[1], 2 for x in [2])
  File "<stdin>", line 1
    f(*[1], 2 for x in [2])

And this was fixed before some prerelease of Python 3.5:

Python 3.5.0a4+ (default:a3f2b171b765, May 19 2015, 16:14:41) 
[GCC 4.9.2] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> f(*[1], 2 for x in [2])
  File "<stdin>", line 1
SyntaxError: Generator expression must be parenthesized if not sole argument

However, the parenthesized generator expression, it works in Python 3.5, but it does not work not in Python 3.4:

f(*[1], (2 for x in [2]))

And this is the clue. In Python 3.5 the *splatting is generalized; you can use it anywhere in a function call:

>>> print(*range(5), 42)
0 1 2 3 4 42

So the actual bug (generator working with *star without parentheses) was indeed fixed in Python 3.5, and the bug could be found in that what changed between Python 3.4 and 3.5

Comments

[J0HN]

Not an exact duplicate, but quite similar: stackoverflow.com/questions/12720450/…. TL;DR seems like it's an implementation detail - it just works like that.

[Bakuriu]

Note: case 2 should work in python 3.5+ (due to the PEP 448)

[Antti Haapala -- Слава Україні]

Python 3.5 is out, and it now tells that the case 3 (actually also the case 4) has been fixed. What's new in Python 3.5

[viraptor]

It's not fixed in 3.5 - just put parens around the generator and the behaviour is the same.

[Antti Haapala -- Слава Україні]

@viraptor good point, in 3.4 the parenthesized expression gives an error

[viraptor]

huh? Running on 3.4.3: f(*[1], 1 for x in [1]) => (<generator object <genexpr> at 0x7fa56c889288>, 1)

[Antti Haapala -- Слава Україні]

@viraptor f(*[1], (1 for x in [1])) is syntax error on Python 3.4. It is valid in Python 3.5.

[Nick Sweeting]

I would gild this answer if I could, thanks for including the relevant C source!