BINARY_CHECKSUM - different result depending on number of rows
I wonder why the BINARY_CHECKSUM
function returns different result for the same:
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL),
(3, 1, 2)) s(id,a,b);
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b);
Ouput:
+-----+----+------+-------------+
| id | a | b | bc |
+-----+----+------+-------------+
| 1 | | 100 | -109 |
| 2 | | | -2147483640 |
| 3 | 1 | 2 | 18 |
+-----+----+------+-------------+
-- -109 vs 100
+-----+----+------+------------+
| id | a | b | bc |
+-----+----+------+------------+
| 1 | | 100 | 100 |
| 2 | | | 2147483647 |
+-----+----+------+------------+
And for second sample I get what I would anticipate:
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, 1, 100),
(2, 3, 4),
(3,1,1)) s(id,a,b);
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, 1, 100),
(2, 3, 4)) s(id,a,b);
Ouptut for both first two rows:
+-----+----+------+-----+
| id | a | b | bc |
+-----+----+------+-----+
| 1 | 1 | 100 | 116 |
| 2 | 3 | 4 | 52 |
+-----+----+------+-----+
db<>fiddle demo
It has strange consequences when I want to compare two tables/queries:
WITH t AS (
SELECT 1 AS id, NULL AS a, 100 b
UNION ALL SELECT 2, NULL, NULL
UNION ALL SELECT 3, 1, 2 -- comment this out
), s AS (
SELECT 1 AS id ,100 AS a, NULL as b
UNION ALL SELECT 2, NULL, NULL
UNION ALL SELECT 3, 2, 1 -- comment this out
)
SELECT t.*,s.*
,BINARY_CHECKSUM(t.a, t.b) AS bc_t, BINARY_CHECKSUM(s.a, s.b) AS bc_s
FROM t
JOIN s
ON s.id = t.id
WHERE BINARY_CHECKSUM(t.a, t.b) = BINARY_CHECKSUM(s.a, s.b);
db<>fiddle demo2
For 3 rows I get single result:
+-----+----+----+-----+----+----+--------------+-------------+
| id | a | b | id | a | b | bc_t | bc_s |
+-----+----+----+-----+----+----+--------------+-------------+
| 2 | | | 2 | | | -2147483640 | -2147483640 |
+-----+----+----+-----+----+----+--------------+-------------+
but for 2 rows I get also id = 1:
+-----+----+------+-----+------+----+-------------+------------+
| id | a | b | id | a | b | bc_t | bc_s |
+-----+----+------+-----+------+----+-------------+------------+
| 1 | | 100 | 1 | 100 | | 100 | 100 |
| 2 | | | 2 | | | 2147483647 | 2147483647 |
+-----+----+------+-----+------+----+-------------+------------+
Remarks:
I am not searching for alternatives like(HASH_BYTES/MD5/CHECKSUM)
I am aware that
BINARY_CHECKSUM
could lead to collisions(two different calls produce the same output) here scenario is a bit different
For this definition, we say that null values, of a specified type,
compare as equal values. If at least one of the values in the
expression list changes, the expression checksum can also change.
However, this is not guaranteed. Therefore, to detect whether values
have changed, we recommend use of BINARY_CHECKSUM only if your
application can tolerate an occasional missed change.
It is strange for me that hash function returns different result for the same input arguments.
Is this behaviour by design or it is some kind of glitch?
EDIT:
As @scsimon
points out it works for materialized tables but not for cte.
db<>fiddle actual table
Metadata for cte:
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL),
(3, 1, 2)) s(id,a,b)', NULL,0);
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b)', NULL,0)
-- working workaround
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, cast(NULL as int), 100),
(2, NULL, NULL)) s(id,a,b)', NULL,0)
For all cases all columns are INT
but with explicit CAST
it behaves as it should.
db<>fidde metadata
sql sql-server tsql
add a comment |
I wonder why the BINARY_CHECKSUM
function returns different result for the same:
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL),
(3, 1, 2)) s(id,a,b);
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b);
Ouput:
+-----+----+------+-------------+
| id | a | b | bc |
+-----+----+------+-------------+
| 1 | | 100 | -109 |
| 2 | | | -2147483640 |
| 3 | 1 | 2 | 18 |
+-----+----+------+-------------+
-- -109 vs 100
+-----+----+------+------------+
| id | a | b | bc |
+-----+----+------+------------+
| 1 | | 100 | 100 |
| 2 | | | 2147483647 |
+-----+----+------+------------+
And for second sample I get what I would anticipate:
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, 1, 100),
(2, 3, 4),
(3,1,1)) s(id,a,b);
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, 1, 100),
(2, 3, 4)) s(id,a,b);
Ouptut for both first two rows:
+-----+----+------+-----+
| id | a | b | bc |
+-----+----+------+-----+
| 1 | 1 | 100 | 116 |
| 2 | 3 | 4 | 52 |
+-----+----+------+-----+
db<>fiddle demo
It has strange consequences when I want to compare two tables/queries:
WITH t AS (
SELECT 1 AS id, NULL AS a, 100 b
UNION ALL SELECT 2, NULL, NULL
UNION ALL SELECT 3, 1, 2 -- comment this out
), s AS (
SELECT 1 AS id ,100 AS a, NULL as b
UNION ALL SELECT 2, NULL, NULL
UNION ALL SELECT 3, 2, 1 -- comment this out
)
SELECT t.*,s.*
,BINARY_CHECKSUM(t.a, t.b) AS bc_t, BINARY_CHECKSUM(s.a, s.b) AS bc_s
FROM t
JOIN s
ON s.id = t.id
WHERE BINARY_CHECKSUM(t.a, t.b) = BINARY_CHECKSUM(s.a, s.b);
db<>fiddle demo2
For 3 rows I get single result:
+-----+----+----+-----+----+----+--------------+-------------+
| id | a | b | id | a | b | bc_t | bc_s |
+-----+----+----+-----+----+----+--------------+-------------+
| 2 | | | 2 | | | -2147483640 | -2147483640 |
+-----+----+----+-----+----+----+--------------+-------------+
but for 2 rows I get also id = 1:
+-----+----+------+-----+------+----+-------------+------------+
| id | a | b | id | a | b | bc_t | bc_s |
+-----+----+------+-----+------+----+-------------+------------+
| 1 | | 100 | 1 | 100 | | 100 | 100 |
| 2 | | | 2 | | | 2147483647 | 2147483647 |
+-----+----+------+-----+------+----+-------------+------------+
Remarks:
I am not searching for alternatives like(HASH_BYTES/MD5/CHECKSUM)
I am aware that
BINARY_CHECKSUM
could lead to collisions(two different calls produce the same output) here scenario is a bit different
For this definition, we say that null values, of a specified type,
compare as equal values. If at least one of the values in the
expression list changes, the expression checksum can also change.
However, this is not guaranteed. Therefore, to detect whether values
have changed, we recommend use of BINARY_CHECKSUM only if your
application can tolerate an occasional missed change.
It is strange for me that hash function returns different result for the same input arguments.
Is this behaviour by design or it is some kind of glitch?
EDIT:
As @scsimon
points out it works for materialized tables but not for cte.
db<>fiddle actual table
Metadata for cte:
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL),
(3, 1, 2)) s(id,a,b)', NULL,0);
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b)', NULL,0)
-- working workaround
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, cast(NULL as int), 100),
(2, NULL, NULL)) s(id,a,b)', NULL,0)
For all cases all columns are INT
but with explicit CAST
it behaves as it should.
db<>fidde metadata
sql sql-server tsql
add a comment |
I wonder why the BINARY_CHECKSUM
function returns different result for the same:
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL),
(3, 1, 2)) s(id,a,b);
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b);
Ouput:
+-----+----+------+-------------+
| id | a | b | bc |
+-----+----+------+-------------+
| 1 | | 100 | -109 |
| 2 | | | -2147483640 |
| 3 | 1 | 2 | 18 |
+-----+----+------+-------------+
-- -109 vs 100
+-----+----+------+------------+
| id | a | b | bc |
+-----+----+------+------------+
| 1 | | 100 | 100 |
| 2 | | | 2147483647 |
+-----+----+------+------------+
And for second sample I get what I would anticipate:
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, 1, 100),
(2, 3, 4),
(3,1,1)) s(id,a,b);
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, 1, 100),
(2, 3, 4)) s(id,a,b);
Ouptut for both first two rows:
+-----+----+------+-----+
| id | a | b | bc |
+-----+----+------+-----+
| 1 | 1 | 100 | 116 |
| 2 | 3 | 4 | 52 |
+-----+----+------+-----+
db<>fiddle demo
It has strange consequences when I want to compare two tables/queries:
WITH t AS (
SELECT 1 AS id, NULL AS a, 100 b
UNION ALL SELECT 2, NULL, NULL
UNION ALL SELECT 3, 1, 2 -- comment this out
), s AS (
SELECT 1 AS id ,100 AS a, NULL as b
UNION ALL SELECT 2, NULL, NULL
UNION ALL SELECT 3, 2, 1 -- comment this out
)
SELECT t.*,s.*
,BINARY_CHECKSUM(t.a, t.b) AS bc_t, BINARY_CHECKSUM(s.a, s.b) AS bc_s
FROM t
JOIN s
ON s.id = t.id
WHERE BINARY_CHECKSUM(t.a, t.b) = BINARY_CHECKSUM(s.a, s.b);
db<>fiddle demo2
For 3 rows I get single result:
+-----+----+----+-----+----+----+--------------+-------------+
| id | a | b | id | a | b | bc_t | bc_s |
+-----+----+----+-----+----+----+--------------+-------------+
| 2 | | | 2 | | | -2147483640 | -2147483640 |
+-----+----+----+-----+----+----+--------------+-------------+
but for 2 rows I get also id = 1:
+-----+----+------+-----+------+----+-------------+------------+
| id | a | b | id | a | b | bc_t | bc_s |
+-----+----+------+-----+------+----+-------------+------------+
| 1 | | 100 | 1 | 100 | | 100 | 100 |
| 2 | | | 2 | | | 2147483647 | 2147483647 |
+-----+----+------+-----+------+----+-------------+------------+
Remarks:
I am not searching for alternatives like(HASH_BYTES/MD5/CHECKSUM)
I am aware that
BINARY_CHECKSUM
could lead to collisions(two different calls produce the same output) here scenario is a bit different
For this definition, we say that null values, of a specified type,
compare as equal values. If at least one of the values in the
expression list changes, the expression checksum can also change.
However, this is not guaranteed. Therefore, to detect whether values
have changed, we recommend use of BINARY_CHECKSUM only if your
application can tolerate an occasional missed change.
It is strange for me that hash function returns different result for the same input arguments.
Is this behaviour by design or it is some kind of glitch?
EDIT:
As @scsimon
points out it works for materialized tables but not for cte.
db<>fiddle actual table
Metadata for cte:
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL),
(3, 1, 2)) s(id,a,b)', NULL,0);
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b)', NULL,0)
-- working workaround
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, cast(NULL as int), 100),
(2, NULL, NULL)) s(id,a,b)', NULL,0)
For all cases all columns are INT
but with explicit CAST
it behaves as it should.
db<>fidde metadata
sql sql-server tsql
I wonder why the BINARY_CHECKSUM
function returns different result for the same:
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL),
(3, 1, 2)) s(id,a,b);
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b);
Ouput:
+-----+----+------+-------------+
| id | a | b | bc |
+-----+----+------+-------------+
| 1 | | 100 | -109 |
| 2 | | | -2147483640 |
| 3 | 1 | 2 | 18 |
+-----+----+------+-------------+
-- -109 vs 100
+-----+----+------+------------+
| id | a | b | bc |
+-----+----+------+------------+
| 1 | | 100 | 100 |
| 2 | | | 2147483647 |
+-----+----+------+------------+
And for second sample I get what I would anticipate:
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, 1, 100),
(2, 3, 4),
(3,1,1)) s(id,a,b);
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, 1, 100),
(2, 3, 4)) s(id,a,b);
Ouptut for both first two rows:
+-----+----+------+-----+
| id | a | b | bc |
+-----+----+------+-----+
| 1 | 1 | 100 | 116 |
| 2 | 3 | 4 | 52 |
+-----+----+------+-----+
db<>fiddle demo
It has strange consequences when I want to compare two tables/queries:
WITH t AS (
SELECT 1 AS id, NULL AS a, 100 b
UNION ALL SELECT 2, NULL, NULL
UNION ALL SELECT 3, 1, 2 -- comment this out
), s AS (
SELECT 1 AS id ,100 AS a, NULL as b
UNION ALL SELECT 2, NULL, NULL
UNION ALL SELECT 3, 2, 1 -- comment this out
)
SELECT t.*,s.*
,BINARY_CHECKSUM(t.a, t.b) AS bc_t, BINARY_CHECKSUM(s.a, s.b) AS bc_s
FROM t
JOIN s
ON s.id = t.id
WHERE BINARY_CHECKSUM(t.a, t.b) = BINARY_CHECKSUM(s.a, s.b);
db<>fiddle demo2
For 3 rows I get single result:
+-----+----+----+-----+----+----+--------------+-------------+
| id | a | b | id | a | b | bc_t | bc_s |
+-----+----+----+-----+----+----+--------------+-------------+
| 2 | | | 2 | | | -2147483640 | -2147483640 |
+-----+----+----+-----+----+----+--------------+-------------+
but for 2 rows I get also id = 1:
+-----+----+------+-----+------+----+-------------+------------+
| id | a | b | id | a | b | bc_t | bc_s |
+-----+----+------+-----+------+----+-------------+------------+
| 1 | | 100 | 1 | 100 | | 100 | 100 |
| 2 | | | 2 | | | 2147483647 | 2147483647 |
+-----+----+------+-----+------+----+-------------+------------+
Remarks:
I am not searching for alternatives like(HASH_BYTES/MD5/CHECKSUM)
I am aware that
BINARY_CHECKSUM
could lead to collisions(two different calls produce the same output) here scenario is a bit different
For this definition, we say that null values, of a specified type,
compare as equal values. If at least one of the values in the
expression list changes, the expression checksum can also change.
However, this is not guaranteed. Therefore, to detect whether values
have changed, we recommend use of BINARY_CHECKSUM only if your
application can tolerate an occasional missed change.
It is strange for me that hash function returns different result for the same input arguments.
Is this behaviour by design or it is some kind of glitch?
EDIT:
As @scsimon
points out it works for materialized tables but not for cte.
db<>fiddle actual table
Metadata for cte:
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL),
(3, 1, 2)) s(id,a,b)', NULL,0);
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b)', NULL,0)
-- working workaround
SELECT name, system_type_name
FROM sys.dm_exec_describe_first_result_set('
SELECT *
FROM (VALUES(1, cast(NULL as int), 100),
(2, NULL, NULL)) s(id,a,b)', NULL,0)
For all cases all columns are INT
but with explicit CAST
it behaves as it should.
db<>fidde metadata
sql sql-server tsql
sql sql-server tsql
edited 27 mins ago
Moira
5,25321837
5,25321837
asked 2 hours ago
Lukasz Szozda
78.7k1061104
78.7k1061104
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
For the literal NULL
without the CAST
(and without any typed values in the column) it entirely ignores it and just gives you the same result as BINARY_CHECKSUM(b)
.
This seems to happen very early on. The initial tree representation output from
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b)
OPTION (RECOMPILE, QUERYTRACEON 8605, QUERYTRACEON 3604);
Already shows that it has decided to just use one column as input to the function
ScaOp_Intrinsic binary_checksum
ScaOp_Identifier COL: Union1008
This compares with the following output for your first query
ScaOp_Intrinsic binary_checksum
ScaOp_Identifier COL: Union1011
ScaOp_Identifier COL: Union1010
If you try and get the BINARY_CHECKSUM
with
SELECT *, BINARY_CHECKSUM(a) AS bc
FROM (VALUES(1, NULL, 100)) s(id,a,b)
It gives the error
Msg 8184, Level 16, State 1, Line 8 Error in binarychecksum. There are
no comparable columns in the binarychecksum input.
This is not the only place where an untyped NULL
constant is treated differently from an explicitly typed one.
Another case is
SELECT COALESCE(CAST(NULL AS INT),CAST(NULL AS INT))
vs
SELECT COALESCE(NULL,NULL)
I'd err on the side of "glitch" in this case rather than "by design" though as the columns from the derived table are supposed to be int
before they get to the checksum function.
SELECT COALESCE(a,b)
FROM (VALUES(NULL, NULL)) s(a,b)
Does work as expected without this glitch.
1
Good catch withCOALESCE
– Lukasz Szozda
1 hour ago
add a comment |
This has nothing to do with the number of rows. It is because the values in one of the columns of the 2-row version are always NULL
. The default type of NULL
is int
and the default type of a numeric constant (of this length) is int
, so these should be comparable. But from a values()
derived table, these are (apparently) not exactly the same type.
In particular, a column with only typeless NULL
s from a derived table is not comparable, so it is excluded from the binary checksum calculation. This does not occur in a real table, because all columns have types.
The rest of the answer illustrates what is happening.
The code behaves as expected with type conversions:
SELECT *, BINARY_CHECKSUM(a, b) AS bc
FROM (VALUES(1, cast(NULL as int), 100),
(2, NULL, NULL)
) s(id,a,b);
Here is a db<>fiddle.
Actually creating tables with the values suggests that columns with only NULL
values have exactly the same type as columns with explicit numbers. That suggests that the original code should work. But an explicit cast also fixes the problem. Very strange.
This is really, really strange. Consider the following:
select v.*, checksum(a, b), checksum(c,b)
FROM (VALUES(1, NULL, 100, NULL),
(2, 1, 2, 1.0)
) v(id, a, b, c);
The change in type for "d" affects the binary_checksum()
for the second row, but not for the first.
This is my conclusion. When all the values in a column are binary, then binary_checksum()
is aware of this and the column is in the category of "noncomparable data type". The checksum is then based on the remaining columns.
You can validate this by seeing the error when you run:
select v.*, binary_checksum(a)
FROM (VALUES(1, NULL, 100, NULL),
(2, NULL, 2, 1.0)
) v( id,a, b, c);
It complains:
Argument data type NULL is invalid for argument 1 of checksum function.
Ironically, this is not a problem if you save the results into a table and use binary_checksum()
. The issue appears to be some interaction with values()
and data types -- but something that is not immediately obvious in the information_schema.columns
table.
The happyish news is that the code should work on tables, even if it does not work on values()
generated derived tables -- as this SQL Fiddle demonstrates.
I also learned that a column filled with NULL
s really is typeless. The assignment of the int
data type in a select into
seems to happen when the table is being defined. The "typeless" type is converted to an int
.
In SQL Server default type for NULL is INT. It is a good observation
– Lukasz Szozda
1 hour ago
1
This doesn't happen when you use an actual table oddly with int datatypes so +1
– scsimon
1 hour ago
3
Datatypes for all cases demo As I thouht all ints - so it looks like a glitch for me
– Lukasz Szozda
1 hour ago
1
@scsimon . . . I am perplexed. According to the information schema tables, the types are exactly the same for allNULL
values and when there is a number constant. However, thecast()
works in this example.
– Gordon Linoff
1 hour ago
@GordonLinoff Thank you for your insight Gordon. I will definitely return to this in 2019.
– Lukasz Szozda
1 hour ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53988972%2fbinary-checksum-different-result-depending-on-number-of-rows%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
For the literal NULL
without the CAST
(and without any typed values in the column) it entirely ignores it and just gives you the same result as BINARY_CHECKSUM(b)
.
This seems to happen very early on. The initial tree representation output from
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b)
OPTION (RECOMPILE, QUERYTRACEON 8605, QUERYTRACEON 3604);
Already shows that it has decided to just use one column as input to the function
ScaOp_Intrinsic binary_checksum
ScaOp_Identifier COL: Union1008
This compares with the following output for your first query
ScaOp_Intrinsic binary_checksum
ScaOp_Identifier COL: Union1011
ScaOp_Identifier COL: Union1010
If you try and get the BINARY_CHECKSUM
with
SELECT *, BINARY_CHECKSUM(a) AS bc
FROM (VALUES(1, NULL, 100)) s(id,a,b)
It gives the error
Msg 8184, Level 16, State 1, Line 8 Error in binarychecksum. There are
no comparable columns in the binarychecksum input.
This is not the only place where an untyped NULL
constant is treated differently from an explicitly typed one.
Another case is
SELECT COALESCE(CAST(NULL AS INT),CAST(NULL AS INT))
vs
SELECT COALESCE(NULL,NULL)
I'd err on the side of "glitch" in this case rather than "by design" though as the columns from the derived table are supposed to be int
before they get to the checksum function.
SELECT COALESCE(a,b)
FROM (VALUES(NULL, NULL)) s(a,b)
Does work as expected without this glitch.
1
Good catch withCOALESCE
– Lukasz Szozda
1 hour ago
add a comment |
For the literal NULL
without the CAST
(and without any typed values in the column) it entirely ignores it and just gives you the same result as BINARY_CHECKSUM(b)
.
This seems to happen very early on. The initial tree representation output from
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b)
OPTION (RECOMPILE, QUERYTRACEON 8605, QUERYTRACEON 3604);
Already shows that it has decided to just use one column as input to the function
ScaOp_Intrinsic binary_checksum
ScaOp_Identifier COL: Union1008
This compares with the following output for your first query
ScaOp_Intrinsic binary_checksum
ScaOp_Identifier COL: Union1011
ScaOp_Identifier COL: Union1010
If you try and get the BINARY_CHECKSUM
with
SELECT *, BINARY_CHECKSUM(a) AS bc
FROM (VALUES(1, NULL, 100)) s(id,a,b)
It gives the error
Msg 8184, Level 16, State 1, Line 8 Error in binarychecksum. There are
no comparable columns in the binarychecksum input.
This is not the only place where an untyped NULL
constant is treated differently from an explicitly typed one.
Another case is
SELECT COALESCE(CAST(NULL AS INT),CAST(NULL AS INT))
vs
SELECT COALESCE(NULL,NULL)
I'd err on the side of "glitch" in this case rather than "by design" though as the columns from the derived table are supposed to be int
before they get to the checksum function.
SELECT COALESCE(a,b)
FROM (VALUES(NULL, NULL)) s(a,b)
Does work as expected without this glitch.
1
Good catch withCOALESCE
– Lukasz Szozda
1 hour ago
add a comment |
For the literal NULL
without the CAST
(and without any typed values in the column) it entirely ignores it and just gives you the same result as BINARY_CHECKSUM(b)
.
This seems to happen very early on. The initial tree representation output from
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b)
OPTION (RECOMPILE, QUERYTRACEON 8605, QUERYTRACEON 3604);
Already shows that it has decided to just use one column as input to the function
ScaOp_Intrinsic binary_checksum
ScaOp_Identifier COL: Union1008
This compares with the following output for your first query
ScaOp_Intrinsic binary_checksum
ScaOp_Identifier COL: Union1011
ScaOp_Identifier COL: Union1010
If you try and get the BINARY_CHECKSUM
with
SELECT *, BINARY_CHECKSUM(a) AS bc
FROM (VALUES(1, NULL, 100)) s(id,a,b)
It gives the error
Msg 8184, Level 16, State 1, Line 8 Error in binarychecksum. There are
no comparable columns in the binarychecksum input.
This is not the only place where an untyped NULL
constant is treated differently from an explicitly typed one.
Another case is
SELECT COALESCE(CAST(NULL AS INT),CAST(NULL AS INT))
vs
SELECT COALESCE(NULL,NULL)
I'd err on the side of "glitch" in this case rather than "by design" though as the columns from the derived table are supposed to be int
before they get to the checksum function.
SELECT COALESCE(a,b)
FROM (VALUES(NULL, NULL)) s(a,b)
Does work as expected without this glitch.
For the literal NULL
without the CAST
(and without any typed values in the column) it entirely ignores it and just gives you the same result as BINARY_CHECKSUM(b)
.
This seems to happen very early on. The initial tree representation output from
SELECT *, BINARY_CHECKSUM(a,b) AS bc
FROM (VALUES(1, NULL, 100),
(2, NULL, NULL)) s(id,a,b)
OPTION (RECOMPILE, QUERYTRACEON 8605, QUERYTRACEON 3604);
Already shows that it has decided to just use one column as input to the function
ScaOp_Intrinsic binary_checksum
ScaOp_Identifier COL: Union1008
This compares with the following output for your first query
ScaOp_Intrinsic binary_checksum
ScaOp_Identifier COL: Union1011
ScaOp_Identifier COL: Union1010
If you try and get the BINARY_CHECKSUM
with
SELECT *, BINARY_CHECKSUM(a) AS bc
FROM (VALUES(1, NULL, 100)) s(id,a,b)
It gives the error
Msg 8184, Level 16, State 1, Line 8 Error in binarychecksum. There are
no comparable columns in the binarychecksum input.
This is not the only place where an untyped NULL
constant is treated differently from an explicitly typed one.
Another case is
SELECT COALESCE(CAST(NULL AS INT),CAST(NULL AS INT))
vs
SELECT COALESCE(NULL,NULL)
I'd err on the side of "glitch" in this case rather than "by design" though as the columns from the derived table are supposed to be int
before they get to the checksum function.
SELECT COALESCE(a,b)
FROM (VALUES(NULL, NULL)) s(a,b)
Does work as expected without this glitch.
edited 1 hour ago
answered 1 hour ago
Martin Smith
341k58572681
341k58572681
1
Good catch withCOALESCE
– Lukasz Szozda
1 hour ago
add a comment |
1
Good catch withCOALESCE
– Lukasz Szozda
1 hour ago
1
1
Good catch with
COALESCE
– Lukasz Szozda
1 hour ago
Good catch with
COALESCE
– Lukasz Szozda
1 hour ago
add a comment |
This has nothing to do with the number of rows. It is because the values in one of the columns of the 2-row version are always NULL
. The default type of NULL
is int
and the default type of a numeric constant (of this length) is int
, so these should be comparable. But from a values()
derived table, these are (apparently) not exactly the same type.
In particular, a column with only typeless NULL
s from a derived table is not comparable, so it is excluded from the binary checksum calculation. This does not occur in a real table, because all columns have types.
The rest of the answer illustrates what is happening.
The code behaves as expected with type conversions:
SELECT *, BINARY_CHECKSUM(a, b) AS bc
FROM (VALUES(1, cast(NULL as int), 100),
(2, NULL, NULL)
) s(id,a,b);
Here is a db<>fiddle.
Actually creating tables with the values suggests that columns with only NULL
values have exactly the same type as columns with explicit numbers. That suggests that the original code should work. But an explicit cast also fixes the problem. Very strange.
This is really, really strange. Consider the following:
select v.*, checksum(a, b), checksum(c,b)
FROM (VALUES(1, NULL, 100, NULL),
(2, 1, 2, 1.0)
) v(id, a, b, c);
The change in type for "d" affects the binary_checksum()
for the second row, but not for the first.
This is my conclusion. When all the values in a column are binary, then binary_checksum()
is aware of this and the column is in the category of "noncomparable data type". The checksum is then based on the remaining columns.
You can validate this by seeing the error when you run:
select v.*, binary_checksum(a)
FROM (VALUES(1, NULL, 100, NULL),
(2, NULL, 2, 1.0)
) v( id,a, b, c);
It complains:
Argument data type NULL is invalid for argument 1 of checksum function.
Ironically, this is not a problem if you save the results into a table and use binary_checksum()
. The issue appears to be some interaction with values()
and data types -- but something that is not immediately obvious in the information_schema.columns
table.
The happyish news is that the code should work on tables, even if it does not work on values()
generated derived tables -- as this SQL Fiddle demonstrates.
I also learned that a column filled with NULL
s really is typeless. The assignment of the int
data type in a select into
seems to happen when the table is being defined. The "typeless" type is converted to an int
.
In SQL Server default type for NULL is INT. It is a good observation
– Lukasz Szozda
1 hour ago
1
This doesn't happen when you use an actual table oddly with int datatypes so +1
– scsimon
1 hour ago
3
Datatypes for all cases demo As I thouht all ints - so it looks like a glitch for me
– Lukasz Szozda
1 hour ago
1
@scsimon . . . I am perplexed. According to the information schema tables, the types are exactly the same for allNULL
values and when there is a number constant. However, thecast()
works in this example.
– Gordon Linoff
1 hour ago
@GordonLinoff Thank you for your insight Gordon. I will definitely return to this in 2019.
– Lukasz Szozda
1 hour ago
add a comment |
This has nothing to do with the number of rows. It is because the values in one of the columns of the 2-row version are always NULL
. The default type of NULL
is int
and the default type of a numeric constant (of this length) is int
, so these should be comparable. But from a values()
derived table, these are (apparently) not exactly the same type.
In particular, a column with only typeless NULL
s from a derived table is not comparable, so it is excluded from the binary checksum calculation. This does not occur in a real table, because all columns have types.
The rest of the answer illustrates what is happening.
The code behaves as expected with type conversions:
SELECT *, BINARY_CHECKSUM(a, b) AS bc
FROM (VALUES(1, cast(NULL as int), 100),
(2, NULL, NULL)
) s(id,a,b);
Here is a db<>fiddle.
Actually creating tables with the values suggests that columns with only NULL
values have exactly the same type as columns with explicit numbers. That suggests that the original code should work. But an explicit cast also fixes the problem. Very strange.
This is really, really strange. Consider the following:
select v.*, checksum(a, b), checksum(c,b)
FROM (VALUES(1, NULL, 100, NULL),
(2, 1, 2, 1.0)
) v(id, a, b, c);
The change in type for "d" affects the binary_checksum()
for the second row, but not for the first.
This is my conclusion. When all the values in a column are binary, then binary_checksum()
is aware of this and the column is in the category of "noncomparable data type". The checksum is then based on the remaining columns.
You can validate this by seeing the error when you run:
select v.*, binary_checksum(a)
FROM (VALUES(1, NULL, 100, NULL),
(2, NULL, 2, 1.0)
) v( id,a, b, c);
It complains:
Argument data type NULL is invalid for argument 1 of checksum function.
Ironically, this is not a problem if you save the results into a table and use binary_checksum()
. The issue appears to be some interaction with values()
and data types -- but something that is not immediately obvious in the information_schema.columns
table.
The happyish news is that the code should work on tables, even if it does not work on values()
generated derived tables -- as this SQL Fiddle demonstrates.
I also learned that a column filled with NULL
s really is typeless. The assignment of the int
data type in a select into
seems to happen when the table is being defined. The "typeless" type is converted to an int
.
In SQL Server default type for NULL is INT. It is a good observation
– Lukasz Szozda
1 hour ago
1
This doesn't happen when you use an actual table oddly with int datatypes so +1
– scsimon
1 hour ago
3
Datatypes for all cases demo As I thouht all ints - so it looks like a glitch for me
– Lukasz Szozda
1 hour ago
1
@scsimon . . . I am perplexed. According to the information schema tables, the types are exactly the same for allNULL
values and when there is a number constant. However, thecast()
works in this example.
– Gordon Linoff
1 hour ago
@GordonLinoff Thank you for your insight Gordon. I will definitely return to this in 2019.
– Lukasz Szozda
1 hour ago
add a comment |
This has nothing to do with the number of rows. It is because the values in one of the columns of the 2-row version are always NULL
. The default type of NULL
is int
and the default type of a numeric constant (of this length) is int
, so these should be comparable. But from a values()
derived table, these are (apparently) not exactly the same type.
In particular, a column with only typeless NULL
s from a derived table is not comparable, so it is excluded from the binary checksum calculation. This does not occur in a real table, because all columns have types.
The rest of the answer illustrates what is happening.
The code behaves as expected with type conversions:
SELECT *, BINARY_CHECKSUM(a, b) AS bc
FROM (VALUES(1, cast(NULL as int), 100),
(2, NULL, NULL)
) s(id,a,b);
Here is a db<>fiddle.
Actually creating tables with the values suggests that columns with only NULL
values have exactly the same type as columns with explicit numbers. That suggests that the original code should work. But an explicit cast also fixes the problem. Very strange.
This is really, really strange. Consider the following:
select v.*, checksum(a, b), checksum(c,b)
FROM (VALUES(1, NULL, 100, NULL),
(2, 1, 2, 1.0)
) v(id, a, b, c);
The change in type for "d" affects the binary_checksum()
for the second row, but not for the first.
This is my conclusion. When all the values in a column are binary, then binary_checksum()
is aware of this and the column is in the category of "noncomparable data type". The checksum is then based on the remaining columns.
You can validate this by seeing the error when you run:
select v.*, binary_checksum(a)
FROM (VALUES(1, NULL, 100, NULL),
(2, NULL, 2, 1.0)
) v( id,a, b, c);
It complains:
Argument data type NULL is invalid for argument 1 of checksum function.
Ironically, this is not a problem if you save the results into a table and use binary_checksum()
. The issue appears to be some interaction with values()
and data types -- but something that is not immediately obvious in the information_schema.columns
table.
The happyish news is that the code should work on tables, even if it does not work on values()
generated derived tables -- as this SQL Fiddle demonstrates.
I also learned that a column filled with NULL
s really is typeless. The assignment of the int
data type in a select into
seems to happen when the table is being defined. The "typeless" type is converted to an int
.
This has nothing to do with the number of rows. It is because the values in one of the columns of the 2-row version are always NULL
. The default type of NULL
is int
and the default type of a numeric constant (of this length) is int
, so these should be comparable. But from a values()
derived table, these are (apparently) not exactly the same type.
In particular, a column with only typeless NULL
s from a derived table is not comparable, so it is excluded from the binary checksum calculation. This does not occur in a real table, because all columns have types.
The rest of the answer illustrates what is happening.
The code behaves as expected with type conversions:
SELECT *, BINARY_CHECKSUM(a, b) AS bc
FROM (VALUES(1, cast(NULL as int), 100),
(2, NULL, NULL)
) s(id,a,b);
Here is a db<>fiddle.
Actually creating tables with the values suggests that columns with only NULL
values have exactly the same type as columns with explicit numbers. That suggests that the original code should work. But an explicit cast also fixes the problem. Very strange.
This is really, really strange. Consider the following:
select v.*, checksum(a, b), checksum(c,b)
FROM (VALUES(1, NULL, 100, NULL),
(2, 1, 2, 1.0)
) v(id, a, b, c);
The change in type for "d" affects the binary_checksum()
for the second row, but not for the first.
This is my conclusion. When all the values in a column are binary, then binary_checksum()
is aware of this and the column is in the category of "noncomparable data type". The checksum is then based on the remaining columns.
You can validate this by seeing the error when you run:
select v.*, binary_checksum(a)
FROM (VALUES(1, NULL, 100, NULL),
(2, NULL, 2, 1.0)
) v( id,a, b, c);
It complains:
Argument data type NULL is invalid for argument 1 of checksum function.
Ironically, this is not a problem if you save the results into a table and use binary_checksum()
. The issue appears to be some interaction with values()
and data types -- but something that is not immediately obvious in the information_schema.columns
table.
The happyish news is that the code should work on tables, even if it does not work on values()
generated derived tables -- as this SQL Fiddle demonstrates.
I also learned that a column filled with NULL
s really is typeless. The assignment of the int
data type in a select into
seems to happen when the table is being defined. The "typeless" type is converted to an int
.
edited 1 hour ago
answered 1 hour ago
Gordon Linoff
758k35291399
758k35291399
In SQL Server default type for NULL is INT. It is a good observation
– Lukasz Szozda
1 hour ago
1
This doesn't happen when you use an actual table oddly with int datatypes so +1
– scsimon
1 hour ago
3
Datatypes for all cases demo As I thouht all ints - so it looks like a glitch for me
– Lukasz Szozda
1 hour ago
1
@scsimon . . . I am perplexed. According to the information schema tables, the types are exactly the same for allNULL
values and when there is a number constant. However, thecast()
works in this example.
– Gordon Linoff
1 hour ago
@GordonLinoff Thank you for your insight Gordon. I will definitely return to this in 2019.
– Lukasz Szozda
1 hour ago
add a comment |
In SQL Server default type for NULL is INT. It is a good observation
– Lukasz Szozda
1 hour ago
1
This doesn't happen when you use an actual table oddly with int datatypes so +1
– scsimon
1 hour ago
3
Datatypes for all cases demo As I thouht all ints - so it looks like a glitch for me
– Lukasz Szozda
1 hour ago
1
@scsimon . . . I am perplexed. According to the information schema tables, the types are exactly the same for allNULL
values and when there is a number constant. However, thecast()
works in this example.
– Gordon Linoff
1 hour ago
@GordonLinoff Thank you for your insight Gordon. I will definitely return to this in 2019.
– Lukasz Szozda
1 hour ago
In SQL Server default type for NULL is INT. It is a good observation
– Lukasz Szozda
1 hour ago
In SQL Server default type for NULL is INT. It is a good observation
– Lukasz Szozda
1 hour ago
1
1
This doesn't happen when you use an actual table oddly with int datatypes so +1
– scsimon
1 hour ago
This doesn't happen when you use an actual table oddly with int datatypes so +1
– scsimon
1 hour ago
3
3
Datatypes for all cases demo As I thouht all ints - so it looks like a glitch for me
– Lukasz Szozda
1 hour ago
Datatypes for all cases demo As I thouht all ints - so it looks like a glitch for me
– Lukasz Szozda
1 hour ago
1
1
@scsimon . . . I am perplexed. According to the information schema tables, the types are exactly the same for all
NULL
values and when there is a number constant. However, the cast()
works in this example.– Gordon Linoff
1 hour ago
@scsimon . . . I am perplexed. According to the information schema tables, the types are exactly the same for all
NULL
values and when there is a number constant. However, the cast()
works in this example.– Gordon Linoff
1 hour ago
@GordonLinoff Thank you for your insight Gordon. I will definitely return to this in 2019.
– Lukasz Szozda
1 hour ago
@GordonLinoff Thank you for your insight Gordon. I will definitely return to this in 2019.
– Lukasz Szozda
1 hour ago
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53988972%2fbinary-checksum-different-result-depending-on-number-of-rows%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown