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PyObject_Compare can now return an error. Unfortunately, there are a 

few places where we don't know how to test for them without losing
speed; don't know yet how to handle that.
This commit is contained in:
Guido van Rossum 1997-05-23 00:01:41 +00:00
parent 7e8d26d78c
commit 5b2121b25f
1 changed files with 8 additions and 0 deletions
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8
Objects/dictobject.c
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@ -184,6 +184,7 @@ lookdict(mp, key, hash)
{
return ep;
}
/* XXX What if PyObject_Compare returned an exception? */
/* Derive incr from sum, just to make it more arbitrary. Note that
incr must not be 0, or we will get into an infinite loop.*/
incr = (sum ^ (sum >> 3)) & mask;
@ -209,6 +210,7 @@ lookdict(mp, key, hash)
PyObject_Compare(ep->me_key, key) == 0)) {
return ep;
}
/* XXX What if PyObject_Compare returned an exception? */
/* Cycle through GF(2^n)-{0} */
incr = incr << 1;
if (incr > mask)
@ -738,10 +740,12 @@ characterize(a, b, pval)
if (a->ma_table[i].me_value != NULL) {
PyObject *key = a->ma_table[i].me_key;
PyObject *aval, *bval;
/* XXX What if PyObject_Compare raises an exception? */
if (diff != NULL && PyObject_Compare(key, diff) > 0)
continue;
aval = a->ma_table[i].me_value;
bval = PyDict_GetItem((PyObject *)b, key);
/* XXX What if PyObject_Compare raises an exception? */
if (bval == NULL || PyObject_Compare(aval, bval) != 0)
{
diff = key;
@ -766,9 +770,13 @@ dict_compare(a, b)
return 1; /* b is shorter */
/* Same length -- check all keys */
adiff = characterize(a, b, &aval);
if (PyErr_Occurred())
return -1;
if (adiff == NULL)
return 0; /* a is a subset with the same length */
bdiff = characterize(b, a, &bval);
if (PyErr_Occurred())
return -1;
/* bdiff == NULL would be impossible now */
res = PyObject_Compare(adiff, bdiff);
if (res == 0)