[issue34641] Curiosity: f((a)=1) is not a syntax error -- why?

[Benjamin Peterson]

Benjamin Peterson  added the comment:


New changeset c9a71dd223c4ad200a97aa320aef6bd3d45f8897 by Benjamin Peterson in 
branch 'master':
closes bpo-34641: Further restrict the LHS of keyword argument function call 
syntax. (GH-9212)
https://github.com/python/cpython/commit/c9a71dd223c4ad200a97aa320aef6bd3d45f8897


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resolution:  -> fixed
stage: patch review -> resolved
status: open -> closed

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[issue34641] Curiosity: f((a)=1) is not a syntax error -- why?

[Karthikeyan Singaravelan]

Karthikeyan Singaravelan  added the comment:

Tested the patch.

➜  cpython git:(master) ✗ ./python.exe
Python 3.8.0a0 (heads/master-dirty:731ff68eee, Sep 12 2018, 11:40:16)
[Clang 7.0.2 (clang-700.1.81)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def foo(a): pass
...
>>> foo((a)=1)
  File "", line 1
SyntaxError: keyword can't be an expression

➜  cpython git:(master) ✗ ./python.exe -m unittest -v test.test_syntax
test_assign_call (test.test_syntax.SyntaxTestCase) ... ok
test_assign_del (test.test_syntax.SyntaxTestCase) ... ok
test_bad_outdent (test.test_syntax.SyntaxTestCase) ... ok
test_break_outside_loop (test.test_syntax.SyntaxTestCase) ... ok
test_global_param_err_first (test.test_syntax.SyntaxTestCase) ... ok
test_kwargs_last (test.test_syntax.SyntaxTestCase) ... ok
test_kwargs_last2 (test.test_syntax.SyntaxTestCase) ... ok
test_kwargs_last3 (test.test_syntax.SyntaxTestCase) ... ok
test_no_indent (test.test_syntax.SyntaxTestCase) ... ok
test_nonlocal_param_err_first (test.test_syntax.SyntaxTestCase) ... ok
test_unexpected_indent (test.test_syntax.SyntaxTestCase) ... ok

--
Ran 11 tests in 0.006s

OK



Thanks

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[issue34641] Curiosity: f((a)=1) is not a syntax error -- why?

[Benjamin Peterson]

Change by Benjamin Peterson :


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keywords: +patch
pull_requests: +8644
stage:  -> patch review

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[issue34641] Curiosity: f((a)=1) is not a syntax error -- why?

[Serhiy Storchaka]

Serhiy Storchaka  added the comment:

Could you please create a PR Benjamin? And please keep the comment about 
special casing lambda.

See also issue30858 about the wording of the error message.

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[issue34641] Curiosity: f((a)=1) is not a syntax error -- why?

[Karthikeyan Singaravelan]

Karthikeyan Singaravelan  added the comment:

There seems to be some of the similar cases that have tests added as part of 
c960f26044edaea6669e60859ecf590c63c65e62 and the original error message added 
at 3e0055f8c65c407e74ce476b8e2b1fb889723514.

Existing tests : 

>>> f(x()=2)
Traceback (most recent call last):
SyntaxError: keyword can't be an expression
>>> f(a or b=1)
Traceback (most recent call last):
SyntaxError: keyword can't be an expression
>>> f(x.y=1)
Traceback (most recent call last):
SyntaxError: keyword can't be an expression


Thanks

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[issue34641] Curiosity: f((a)=1) is not a syntax error -- why?

[Serhiy Storchaka]

Change by Serhiy Storchaka :


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nosy: +serhiy.storchaka

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[issue34641] Curiosity: f((a)=1) is not a syntax error -- why?

[Karthikeyan Singaravelan]

Change by Karthikeyan Singaravelan :


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nosy: +xtreak

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[issue34641] Curiosity: f((a)=1) is not a syntax error -- why?

[Benjamin Peterson]

Benjamin Peterson  added the comment:

I expect you'd have to make the check of test nodes in ast.c stricter. Here's a 
slightly gross implementation of that:

diff --git a/Python/ast.c b/Python/ast.c
index 94962e00c7..b7cebf4777 100644
--- a/Python/ast.c
+++ b/Python/ast.c
@@ -2815,15 +2815,22 @@ 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)
+static const int chain[] = {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; chain[i]; i++) {
+if (TYPE(expr_node) != chain[i])
+break;
+if (NCH(expr_node) != 1)
+break;
+expr_node = CHILD(expr_node, 0);
+}
+if (TYPE(expr_node) != NAME) {
+ast_error(c, chch,
+"keyword can't be an expression");
 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.
- */
+}
+e = ast_for_expr(c, chch);
 if (e->kind == Lambda_kind) {
 ast_error(c, chch,
 "lambda cannot contain assignment");

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nosy: +benjamin.peterson

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[issue34641] Curiosity: f((a)=1) is not a syntax error -- why?

[Guido van Rossum]

New submission from Guido van Rossum :

Emily and I just discovered that f((a)=1) is accepted and compiled the same as 
f(a=1). This goes against the intention that keyword arguments have the syntax 
f(NAME=expr).

I suspect that this behavior was introduced at the time we switched from 
generating from the (concrete) parse tree to first converting to an ast and 
then generating code from there. I.e. in the distant past (long before 2.7 
even).

But I still think it ought to be fixed. Thoughts? Anyone have an idea *where* 
to fix it?

--
messages: 325107
nosy: emilyemorehouse, gvanrossum
priority: normal
severity: normal
status: open
title: Curiosity: f((a)=1) is not a syntax error -- why?
versions: Python 3.8

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