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closes bpo-34641: Further restrict the LHS of keyword argument function call syntax. (GH-9212)

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Benjamin Peterson 2018-09-12 17:14:39 -07:00 committed by GitHub
parent 6d9767fb26
commit c9a71dd223
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4 changed files with 50 additions and 14 deletions
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4
Doc/whatsnew/3.8.rst
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@ -94,6 +94,10 @@ Other Language Changes
* Added support of ``\N{name}`` escapes in :mod:`regular expressions <re>`.
(Contributed by Jonathan Eunice and Serhiy Storchaka in :issue:`30688`.)
* The syntax allowed for keyword names in function calls was further
restricted. In particular, ``f((keyword)=arg)`` is no longer allowed. It was
never intended to permit more than a bare name on the left-hand side of a
keyword argument assignment term. See :issue:`34641`.
New Modules
===========

3
Lib/test/test_syntax.py
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@ -269,6 +269,9 @@ SyntaxError: keyword can't be an expression
>>> f(x.y=1)
Traceback (most recent call last):
SyntaxError: keyword can't be an expression
>>> f((x)=2)
Traceback (most recent call last):
SyntaxError: keyword can't be an expression
More set_context():

2
Misc/NEWS.d/next/Core and Builtins/2018-09-11-23-12-33.bpo-34641.gFBCc9.rst Normal file
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@ -0,0 +1,2 @@
Further restrict the syntax of the left-hand side of keyword arguments in
function calls. In particular, ``f((keyword)=arg)`` is now disallowed.

55
Python/ast.c
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@ -2815,29 +2815,56 @@ ast_for_call(struct compiling *c, const node *n, expr_ty func, bool allowgen)
identifier key, tmp;
int k;
/* chch is test, but must be an identifier? */
e = ast_for_expr(c, chch);
if (!e)
return NULL;
/* f(lambda x: x[0] = 3) ends up getting parsed with
* LHS test = lambda x: x[0], and RHS test = 3.
* SF bug 132313 points out that complaining about a keyword
* then is very confusing.
*/
if (e->kind == Lambda_kind) {
// To remain LL(1), the grammar accepts any test (basically, any
// expression) in the keyword slot of a call site. So, we need
// to manually enforce that the keyword is a NAME here.
static const int name_tree[] = {
test,
or_test,
and_test,
not_test,
comparison,
expr,
xor_expr,
and_expr,
shift_expr,
arith_expr,
term,
factor,
power,
atom_expr,
atom,
0,
};
node *expr_node = chch;
for (int i = 0; name_tree[i]; i++) {
if (TYPE(expr_node) != name_tree[i])
break;
if (NCH(expr_node) != 1)
break;
expr_node = CHILD(expr_node, 0);
}
if (TYPE(expr_node) == lambdef) {
// f(lambda x: x[0] = 3) ends up getting parsed with LHS
// test = lambda x: x[0], and RHS test = 3. Issue #132313
// points out that complaining about a keyword then is very
// confusing.
ast_error(c, chch,
"lambda cannot contain assignment");
return NULL;
}
else if (e->kind != Name_kind) {
else if (TYPE(expr_node) != NAME) {
ast_error(c, chch,
"keyword can't be an expression");
"keyword can't be an expression");
return NULL;
}
else if (forbidden_name(c, e->v.Name.id, ch, 1)) {
key = new_identifier(STR(expr_node), c);
if (key == NULL) {
return NULL;
}
if (forbidden_name(c, key, chch, 1)) {
return NULL;
}
key = e->v.Name.id;
for (k = 0; k < nkeywords; k++) {
tmp = ((keyword_ty)asdl_seq_GET(keywords, k))->arg;
if (tmp && !PyUnicode_Compare(tmp, key)) {