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MDEV-10657: incorrect result returned with binary protocol (prepared statements)

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If translation table present when we materialize the derived table then
change it to point to the materialized table.

Added debug info to see really what happens with what derived.
This commit is contained in:
Oleksandr Byelkin
2017-12-20 13:52:27 +01:00
committed by Oleksandr Byelkin
parent 924db8b4ed
commit 462808f3b6
6 changed files with 96 additions and 3 deletions
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9
sql/sql_class.cc
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@ -2192,6 +2192,8 @@ void THD::nocheck_register_item_tree_change(Item **place, Item *old_value,
MEM_ROOT *runtime_memroot)
{
Item_change_record *change;
DBUG_ENTER("THD::nocheck_register_item_tree_change");
DBUG_PRINT("enter", ("Register %p <- %p", old_value, (*place)));
/*
Now we use one node per change, which adds some memory overhead,
but still is rather fast as we use alloc_root for allocations.
@ -2204,12 +2206,13 @@ void THD::nocheck_register_item_tree_change(Item **place, Item *old_value,
OOM, thd->fatal_error() is called by the error handler of the
memroot. Just return.
*/
return;
DBUG_VOID_RETURN;
}
change= new (change_mem) Item_change_record;
change->place= place;
change->old_value= old_value;
change_list.append(change);
DBUG_VOID_RETURN;
}
/**
@ -2250,7 +2253,11 @@ void THD::rollback_item_tree_changes()
DBUG_ENTER("rollback_item_tree_changes");
while ((change= it++))
{
DBUG_PRINT("info", ("revert %p -> %p",
change->old_value, (*change->place)));
*change->place= change->old_value;
}
/* We can forget about changes memory: it's allocated in runtime memroot */
change_list.empty();
DBUG_VOID_RETURN;