or
Paul White
sql-server
In answer to [SQL counting distinct over partition](https://dba.stackexchange.com/q/239788), Erik Darling posted this code to work around for the lack of `COUNT(DISTINCT) OVER ()`:

```
SELECT      *
FROM        #MyTable AS mt
CROSS APPLY (   SELECT COUNT(DISTINCT mt2.Col_B) AS dc
                FROM   #MyTable AS mt2
                WHERE  mt2.Col_A = mt.Col_A
                -- GROUP BY mt2.Col_A 
            ) AS ca;
```

The query uses `CROSS APPLY` (not `OUTER APPLY`) so why is there an **outer** join in the execution plan instead of an **inner** join?

[![enter image description here][1]][1]

Also why does uncommenting the group by clause result in an inner join?

[![enter image description here][2]][2]

I don't think the data is important but copying from that given by kevinwhat on the other question:

```
create table #MyTable (
Col_A varchar(5),
Col_B int
)

insert into #MyTable values ('A',1)
insert into #MyTable values ('A',1)
insert into #MyTable values ('A',2)
insert into #MyTable values ('A',2)
insert into #MyTable values ('A',2)
insert into #MyTable values ('A',3)

insert into #MyTable values ('B',4)
insert into #MyTable values ('B',4)
insert into #MyTable values ('B',5)
```

  [1]: https://i.stack.imgur.com/oMRf9.png
  [2]: https://i.stack.imgur.com/DYdNO.png
Top Answer
Paul White
## Summary

SQL Server uses the correct join (inner or outer) and adds projections where necessary to **honour all the semantics** of the original query when performing [internal translations][1] between *apply* and *join*.

The differences in the plans can all be explained by the [different semantics][2] of aggregates with and without a group by clause in SQL Server.

---

## Details

### Join vs Apply

We will need to be able to distinguish between an *apply* and a *join*:

* **Apply** 

 The inner (lower) input of the *apply* is run for each row of the outer (upper) input, with one or more inner side parameter values provided by the current outer row. The overall result of the *apply* is the combination (union all) of all the rows produced by the parameterized inner side executions. The presence of parameters means *apply* is sometimes referred to as a correlated join.

 An *apply* is always implemented in execution plans by the *Nested Loops* operator. The operator will have an *Outer References* property rather than join predicates. The outer references are the parameters passed from the outer side to the inner side on each iteration of the loop.

* **Join**

 A join evaluates its join predicate at the join operator. The join may generally be implemented by *Hash Match*, *Merge*, or *Nested Loops* operators in SQL Server.

 When *Nested Loops* is chosen, it can be distinguished from an *apply* by the lack of *Outer References* (and usually the presence of a join predicate). The inner input of a *join* never references values from the outer input - the inner side is still executed once for each outer row, but inner side executions do not depend on any values from the current outer row.

For more details see my post [Apply versus Nested Loops Join][1].

>...why is there an **outer** join in the execution plan instead of an **inner** join?

The outer join arises when the optimizer transforms an *apply* to a *join* (using a rule called `ApplyHandler`) to see if it can find a cheaper join-based plan. The join is required to be an outer join for *correctness* when the *apply* contains a *scalar aggregate*. An inner join would not be *guaranteed* to produce the same results as the original *apply* as we will see.

### Scalar and Vector Aggregates

* An aggregate without a corresponding `GROUP BY` clause is a **scalar** aggregate.
* An aggregate with a corresponding `GROUP BY` clause is a **vector** aggregate.

In SQL Server, a *scalar* aggregate will always produce a row, even if it is given no rows to aggregate. For example, the scalar `COUNT` aggregate of no rows is zero. A *vector* `COUNT` aggregate of no rows is the empty set (no rows at all).

The following toy queries illustrate the difference. You can also read more about scalar and vector aggregates in my article [Fun with Scalar and Vector Aggregates][2].

```
-- Produces a single zero value
SELECT COUNT_BIG(*) FROM #MyTable AS MT WHERE 0 = 1;

-- Produces no rows
SELECT COUNT_BIG(*) FROM #MyTable AS MT WHERE 0 = 1 GROUP BY ();
```

Demo:

<>https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=ae2b672d78d5bd51ec8b29abe301e078

### Transforming apply to join

I mentioned before that the join is required to be an outer join for *correctness* when the original *apply* contains a *scalar aggregate*. To show why this is the case in detail, I will use a simplified example of the question query:

```
DECLARE @A table (A integer NULL, B integer NULL);
DECLARE @B table (A integer NULL, B integer NULL);

INSERT @A (A, B) VALUES (1, 1);
INSERT @B (A, B) VALUES (2, 2);

SELECT * FROM @A AS A
CROSS APPLY (SELECT c = COUNT_BIG(*) FROM @B AS B WHERE B.A = A.A) AS CA;
```

The correct result for column `c` is **zero**, because the `COUNT_BIG` is a **scalar** aggregate. When translating this apply query to join form, SQL Server generates an internal alternative that would look similar to the following if it were expressed in T-SQL:

```
SELECT A.*, c = COALESCE(J1.c, 0)
FROM @A AS A
LEFT JOIN
(
    SELECT B.A, c = COUNT_BIG(*) 
    FROM @B AS B
    GROUP BY B.A
) AS J1
    ON J1.A = A.A;
```

To rewrite the apply as an uncorrelated join, we have to introduce a `GROUP BY` in the derived table (otherwise there could be no `A` column to join on). The join has to be an **outer** join so each row from table `@A` continues to produce a row in the output. The left join will produce a `NULL` for column `c` when the join predicate does not evaluate to true. That `NULL` needs to be translated to zero by `COALESCE` to complete a correct transformation from *apply*.

The demo below shows how both outer join and `COALESCE` are required to produce the same results using *join* as the original *apply* query:

<>https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=8302f12448fffe6ed0ce192791cebd4c

### With the `GROUP BY`

>...why does uncommenting the group by clause result in an inner join?

Continuing the simplified example, but adding a `GROUP BY`:

```
DECLARE @A table (A integer NULL, B integer NULL);
DECLARE @B table (A integer NULL, B integer NULL);

INSERT @A (A, B) VALUES (1, 1);
INSERT @B (A, B) VALUES (2, 2);

-- Original
SELECT * FROM @A AS A
CROSS APPLY 
(SELECT c = COUNT_BIG(*) FROM @B AS B WHERE B.A = A.A GROUP BY B.A) AS CA;

```

The `COUNT_BIG` is now a **vector** aggregate, so the correct result for an empty input set is no longer zero, it is **no row at all**. In other words, running the statements above produces no output.

These semantics are much easier to honour when translating from *apply* to *join*,  since `CROSS APPLY` naturally rejects any outer row that generates no inner side rows. We can therefore safely use an inner join now, with no extra expression projection:

```
-- Rewrite
SELECT A.*, J1.c 
FROM @A AS A
JOIN
(
    SELECT B.A, c = COUNT_BIG(*) 
    FROM @B AS B
    GROUP BY B.A
) AS J1
    ON J1.A = A.A;
```

The demo below shows that the inner join rewrite produces the same results as the original apply with vector aggregate:

<>https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=43a1f0fd09c650e201e0441c77847248

The optimizer happens to choose a merge inner join with the small table because it finds a cheap *join* plan quickly (good enough plan found). The cost based optimizer may go on to rewrite the join back to an apply - perhaps finding a cheaper apply plan, as it will here if a loop join or forceseek hint is used - but it is not worth the effort in this case.

### Notes

The simplified examples use different tables with different contents to show the semantic differences more clearly.

One could argue that the optimizer ought to be able to reason about a self-join not being capable generating any mismatched (non-joining) rows, but it does not contain that logic today. Accessing the same table multiple times in a query is not guaranteed to produce the same results in general anyway, depending on isolation level and concurrent activity.

The optimizer worries about these semantics and edge cases so you don't have to.

---

### Bonus: Inner *Apply* Plan

SQL Server **can** produce an inner *apply* plan (not an inner *join* plan!) for the example query, it just chooses not to for cost reasons. The cost of the outer join plan shown in the question is **0.02898** units on my laptop's SQL Server 2017 instance.

You can force an *apply* (correlated join) plan using undocumented and unsupported trace flag 9114 (which disables `ApplyHandler` etc.) just for illustration:

```
SELECT      *
FROM        #MyTable AS mt
CROSS APPLY 
(
    SELECT COUNT_BIG(DISTINCT mt2.Col_B) AS dc
    FROM   #MyTable AS mt2
    WHERE  mt2.Col_A = mt.Col_A 
    --GROUP BY mt2.Col_A
) AS ca
OPTION (QUERYTRACEON 9114);
```

This produces an *apply* nested loops plan with a lazy index spool. The total estimated cost is **0.0463983** (higher than the selected plan):

[![Index Spool apply plan][6]][6]

Note that the execution plan using *apply* nested loops produces correct results using "inner join" semantics regardless of the presence of the `GROUP BY` clause.

In the real world, we would typically have an index to support a seek on the inner side of the *apply* to encourage SQL Server to choose this option naturally, for example:

```
CREATE INDEX i ON #MyTable (Col_A, Col_B);
```

Demo:

<>https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=b314493c20a1b5206391c540400a1dae


  [1]: https://www.sql.kiwi/2019/06/apply-versus-nested-loops-join.html
  [2]: https://www.sql.kiwi/2012/03/fun-with-aggregates.html
  [3]: https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=ae2b672d78d5bd51ec8b29abe301e078
  [4]: https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=8302f12448fffe6ed0ce192791cebd4c
  [5]: https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=43a1f0fd09c650e201e0441c77847248
  [6]: https://i.stack.imgur.com/TCrjK.png
  [7]: https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=b314493c20a1b5206391c540400a1dae
CROSS APPLY produces outer join

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