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Return the actual default partition in dds_qget_partition() whenever possible #1824

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Return the actual default partition in dds_qget_partition() whenever possible #1824

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16 changes: 14 additions & 2 deletions src/core/ddsc/src/dds_qos.c
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Expand Up @@ -618,8 +618,20 @@ bool dds_qget_partition (const dds_qos_t * __restrict qos, uint32_t *n, char ***
*n = qos->partition.n;
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I don't like the behaviour that, for an entity created with an empty set of partitions:

dds_qget_partition (q, &nA, NULL);
dds_qget_partition (q, &nB, &ps);
assert (nA == nB);

will fail.

if (ps)
{
if (qos->partition.n == 0)
*ps = NULL;
if (qos->partition.n == 0) {
/* This is the default partition.
* Only when the argument n is non-null, we'll
* actually return the default partition (i.e.,
* the empty string "").
* */
if (n) {
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Lines 615 & 616 ensure (ps ≠ NULL) ⇒ (n ≠ NULL), so we are certain to never end up in the else case. I think the comment that says "this is the default partition" might better be rephrased as "this implies the default partition", then the remainder of the comment, the if (n) and the else branch can be thrown out.

*n = 1;
*ps = dds_alloc (sizeof (char*));
(*ps)[0] = dds_string_dup ("");
} else {
*ps = NULL;
}
}
else
{
*ps = dds_alloc (sizeof (char*) * qos->partition.n);
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