bpo-37540: vectorcall: keyword names must be strings (GH-14682)

The fact that keyword names are strings is now part of the vectorcall and `METH_FASTCALL` protocols. The biggest concrete change is that `_PyStack_UnpackDict` now checks that and raises `TypeError` if not.

CC @markshannon @vstinner 


https://bugs.python.org/issue37540

1 files changed, 19 insertions, 6 deletions

diff --git a/Objects/call.c b/Objects/call.c
index 7d91789..8a60b3e 100644
--- a/Objects/call.c
+++ b/Objects/call.c
@@ -322,8 +322,7 @@ _PyFunction_Vectorcall(PyObject *func, PyObject* const* stack,
     assert(nargs >= 0);
     assert(kwnames == NULL || PyTuple_CheckExact(kwnames));
     assert((nargs == 0 && nkwargs == 0) || stack != NULL);
-    /* kwnames must only contains str strings, no subclass, and all keys must
-       be unique */
+    /* kwnames must only contain strings and all keys must be unique */
 
     if (co->co_kwonlyargcount == 0 && nkwargs == 0 &&
         (co->co_flags & ~PyCF_MASK) == (CO_OPTIMIZED | CO_NEWLOCALS | CO_NOFREE))
@@ -943,12 +942,12 @@ _PyStack_AsDict(PyObject *const *values, PyObject *kwnames)
    vector; return NULL with exception set on error. Return the keyword names
    tuple in *p_kwnames.
 
-   The newly allocated argument vector supports PY_VECTORCALL_ARGUMENTS_OFFSET.
+   This also checks that all keyword names are strings. If not, a TypeError is
+   raised.
 
-   When done, you must call _PyStack_UnpackDict_Free(stack, nargs, kwnames)
+   The newly allocated argument vector supports PY_VECTORCALL_ARGUMENTS_OFFSET.
 
-   The type of keyword keys is not checked, these checks should be done
-   later (ex: _PyArg_ParseStackAndKeywords). */
+   When done, you must call _PyStack_UnpackDict_Free(stack, nargs, kwnames) */
 static PyObject *const *
 _PyStack_UnpackDict(PyObject *const *args, Py_ssize_t nargs, PyObject *kwargs,
                     PyObject **p_kwnames)
@@ -994,7 +993,9 @@ _PyStack_UnpackDict(PyObject *const *args, Py_ssize_t nargs, PyObject *kwargs,
        called in the performance critical hot code. */
     Py_ssize_t pos = 0, i = 0;
     PyObject *key, *value;
+    unsigned long keys_are_strings = Py_TPFLAGS_UNICODE_SUBCLASS;
     while (PyDict_Next(kwargs, &pos, &key, &value)) {
+        keys_are_strings &= Py_TYPE(key)->tp_flags;
         Py_INCREF(key);
         Py_INCREF(value);
         PyTuple_SET_ITEM(kwnames, i, key);
@@ -1002,6 +1003,18 @@ _PyStack_UnpackDict(PyObject *const *args, Py_ssize_t nargs, PyObject *kwargs,
         i++;
     }
 
+    /* keys_are_strings has the value Py_TPFLAGS_UNICODE_SUBCLASS if that
+     * flag is set for all keys. Otherwise, keys_are_strings equals 0.
+     * We do this check once at the end instead of inside the loop above
+     * because it simplifies the deallocation in the failing case.
+     * It happens to also make the loop above slightly more efficient. */
+    if (!keys_are_strings) {
+        PyErr_SetString(PyExc_TypeError,
+                        "keywords must be strings");
+        _PyStack_UnpackDict_Free(stack, nargs, kwnames);
+        return NULL;
+    }
+
     *p_kwnames = kwnames;
     return stack;
 }