Re: pg_advisory_lock lock FAILURE / What does those numbers mean(process 240828 waits for ExclusiveLock on advisory lock [1167570,16820923,3422556162,1];)?

On Fri, 2019-07-19 at 21:15 +0300, Alexandru Lazarev wrote:
> I receive locking failure on pg_advisory_lock, I do deadlock condition and receive following:
> - - -
> ERROR: deadlock detected
> SQL state: 40P01
> Detail: Process 240828 waits for ExclusiveLock on advisory lock [1167570,16820923,3422556162,1]; blocked by process 243637.
> Process 243637 waits for ExclusiveLock on advisory lock [1167570,16820923,3422556161,1]; blocked by process 240828.
> - - -
> I do from Tx1:
> select pg_advisory_lock(72245317596090369);
> select pg_advisory_lock(72245317596090370);
> and from Tx2:
> select pg_advisory_lock(72245317596090370);
> select pg_advisory_lock(72245317596090369);
>
> where long key is following: 72245317596090369-> HEX 0x0100AABBCC001001
> where 1st byte (highest significance "0x01") is namespace masked with MAC Address " AABBCC001001", but in error i see 4 numbers - what is their meaning?
> I deducted that 2nd ( 16820923 .) HEX 0x100AABB, 1st half of long key) and 3rd is ( 3422556161 -> HEX 0xCC001001, 2nd half of long key)
> but what are 1st ( 1167570 ) and 4th (1) numbers?

See this code in src/backend/utils/adt/lockfuncs.c:

/*
 * Functions for manipulating advisory locks
 *
 * We make use of the locktag fields as follows:
 *
 *  field1: MyDatabaseId ... ensures locks are local to each database
 *  field2: first of 2 int4 keys, or high-order half of an int8 key
 *  field3: second of 2 int4 keys, or low-order half of an int8 key
 *  field4: 1 if using an int8 key, 2 if using 2 int4 keys
 */
#define SET_LOCKTAG_INT64(tag, key64) \
    SET_LOCKTAG_ADVISORY(tag, \
                         MyDatabaseId, \
                         (uint32) ((key64) >> 32), \
                         (uint32) (key64), \
                         1)
#define SET_LOCKTAG_INT32(tag, key1, key2) \
    SET_LOCKTAG_ADVISORY(tag, MyDatabaseId, key1, key2, 2)

Yours,
Laurenz Albe
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