How to find size of index in table-valued function - sql

Question

In article about sys.indexes there is a phrase that this view
Contains a row per index or heap of a tabular object, such as a table,
view, or table-valued function.
I was interested to find a size of such an index.
So I created function with index:
create function fIndexSize()
returns #res table
(
object_id int not null
, name varchar(128) not null
, primary key (object_id)
)
as
begin
insert into #res
select object_id, name
from sys.objects
where object_id > 255
return
end
Here we can see the name of new index:
There is also a record in sys.indexes:
Usually I get sizes of indexes using this query:
select
o.schema_id
, o.object_id
, o.name
, o.type_desc
, sum (a.total_pages) * 8.00 / 1024 / 1024 as TotalSpaceGB
from sys.objects o
inner join sys.indexes i on o.object_id = i.object_id
inner join sys.partitions p on i.object_id = p.object_id and i.index_id = p.index_id
inner join sys.allocation_units a on p.partition_id = a.container_id
where (o.name = 'fIndexSize' or i.name like 'PK__fIndexSi%')
group by o.schema_id, o.object_id, o.name, o.type_desc
But this time nothing was returned.
Can anyone give me advice how to find size of such an index?

Answer 1

Yes you can find the size of this index, but you should consider it's living only for a time of a batch and you should look for it in tempdb (as it is table variable):
create function fIndexSize()
returns #res table
(
object_id_xxxx int not null
, name varchar(128) not null
, primary key (object_id_xxxx)
)
as
begin
insert into #res
select object_id, name
from sys.objects
where object_id > 255
return
end;
go
select i.name,
c.name,
8 * SUM(au.used_pages) as size_kb
from tempdb.sys.indexes i
join tempdb.sys.columns c
on i.object_id = c.object_id
join tempdb.sys.partitions as p
on p.object_id = i.object_id and p.index_id = i.index_id
join tempdb.sys.allocation_units as au
on au.container_id = p.partition_id
where c.name = 'object_id_xxxx'
group by i.name,
c.name
I left the column name here only to show that the index found is what we are looking for, and I chose the column name with xxxx for distinguish it well

Answer 2

The result of a table-valued function is not stored in a permanent table in the database. It is generated on the fly during the query execution.
Yes, you have a row in sys.indexes which tells you index properties, like type (clustered or not), is_primary_key, is_unique, etc.
But, there are no corresponding rows in sys.partitions and in sys.allocation_units. That's why your query returns nothing. If you replace inner joins with left joins, you'd see one row with NULL as TotalSpaceGB.
So, documentation is correct. Documentation doesn't say that table-valued functions will have rows in sys.allocation_units.
Each invocation of the function may return different number of rows. This set of rows doesn't exist before the query runs and it doesn't exist after the query finishes.
Even during the function execution sys.partitions and sys.allocation_units are empty for this index (PK__fIndexSi...).
When I looked at the actual execution plan of the query
select * from fIndexSize()
I could see that optimiser creates a temp table behind the scenes. Well, it has to store the rows somewhere and they are stored in TempDB.
So, you should run your select from sys.allocation_units using tempdb.
At first I used SQL Sentry Plan Explorer to see the name of the temporary table:
Then I ran your query against TempDB: