Re: Default Partition for Range

On Tue, Jun 20, 2017 at 6:57 PM, Dilip Kumar <dilipbalaut(at)gmail(dot)com> wrote:
> This is basically crashing in RelationBuildPartitionDesc, so I think
> we don't have any test case for testing DEFAULT range partition where
> partition key has more than one attribute. So I suggest we can add
> such test case.

Some more comments.

<code>
- bound->datums = (Datum *) palloc0(key->partnatts * sizeof(Datum));
- bound->content = (RangeDatumContent *) palloc0(key->partnatts *
- sizeof(RangeDatumContent));
+ bound->datums = datums ? (Datum *) palloc0(key->partnatts * sizeof(Datum))
+ : NULL;
+ bound->content = (RangeDatumContent *) palloc0(
+ (datums ? key->partnatts : 1) * sizeof(RangeDatumContent));
bound->lower = lower;

i = 0;
+
+ /* If default, datums are NULL */
+ if (datums == NULL)
+ bound->content[i] = RANGE_DATUM_DEFAULT;
</code>

For the default partition we are only setting bound->content[0] to
default, but content for others key
attributes are not initialized. But later in the code, if the content
of the first key is RANGE_DATUM_DEFAULT then it should not access the
next content, but I see there are some exceptions. Which can access
uninitialized value?

for example see below code

<code>
for (i = 0; i < key->partnatts; i++)
{
+ if (rb_content[i] == RANGE_DATUM_DEFAULT) --> why it's going to
content for next parttion attribute, we never initialized that?
+ continue;
+
if (rb_content[i] != RANGE_DATUM_FINITE)
return rb_content[i] == RANGE_DATUM_NEG_INF ? -1 : 1;
</code>

Also
In RelatiobBuildPartitionDesc

<code>
for (j = 0; j < key->partnatts; j++)
{
-- Suppose first is RANGE_DATUM_DEFAULT then we should not check next
because that is never initialized. I think this is the cause of
the crash also what I have reported above.
----
if (rbounds[i]->content[j] == RANGE_DATUM_FINITE)
boundinfo->datums[i][j] =
datumCopy(rbounds[i]->datums[j],
key->parttypbyval[j],
key->parttyplen[j]);
/* Remember, we are storing the tri-state value. */
boundinfo->content[i][j] = rbounds[i]->content[j];
</code>

--
Regards,
Dilip Kumar
EnterpriseDB: http://www.enterprisedb.com