Bytes in Data Page Row

I am reading a book titled "Microsoft SQL Server 2008 Internals". In Chapter
6, which was authored by Kalen Delaney, Kimberly Tripp, and Paul Randall, the
following table and Clustered Index is created:

CREATE TABLE Employee(
  EmployeeID INT NOT NULL IDENTITY,
  LastName NCHAR(30) NOT NULL,
  FirstName NCHAR(29) NOT NULL,
  MiddleInitial NCHAR(1) NULL,
  SSN CHAR(11) NOT NULL,
  OtherColumns CHAR(258) NOT NULL DEFAULT 'Junk')
GO

ALTER TABLE Employee
ADD CONSTRAINT EmployeePK
PRIMARY KEY CLUSTERED (EmployeeID)
GO

It states on page 321 of the aforementioned book, that "the data rows of the
Employee table are exactly 400 bytes".

I am only computing 333 bytes, as per the following:
EmployeeID = 4 bytes
LastName = 30 bytes
FirstName = 29 bytes
MiddleInitial = 1 byte
SSN = 11 bytes
OtherColumns = 258 bytes

Total Bytes = 333 bytes

Can someone explain to me the 67 bytes (400 - 333 = 67) I am unable to
account for?

-- 
Message posted via SQLMonster.com
http://www.sqlmonster.com/Uwe/Forums.aspx/sql-server/200912/1

0
cbrichards
12/4/2009 6:50:02 PM
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On Dec 4, 1:50=A0pm, "cbrichards via SQLMonster.com" <u3288@uwe> wrote:
> I am reading a book titled "Microsoft SQL Server 2008 Internals". In Chap=
ter
> 6, which was authored by Kalen Delaney, Kimberly Tripp, and Paul Randall,=
 the
> following table and Clustered Index is created:
>
> CREATE TABLE Employee(
> =A0 EmployeeID INT NOT NULL IDENTITY,
> =A0 LastName NCHAR(30) NOT NULL,
> =A0 FirstName NCHAR(29) NOT NULL,
> =A0 MiddleInitial NCHAR(1) NULL,
> =A0 SSN CHAR(11) NOT NULL,
> =A0 OtherColumns CHAR(258) NOT NULL DEFAULT 'Junk')
> GO
>
> ALTER TABLE Employee
> ADD CONSTRAINT EmployeePK
> PRIMARY KEY CLUSTERED (EmployeeID)
> GO
>
> It states on page 321 of the aforementioned book, that "the data rows of =
the
> Employee table are exactly 400 bytes".
>
> I am only computing 333 bytes, as per the following:
> EmployeeID =3D 4 bytes
> LastName =3D 30 bytes
> FirstName =3D 29 bytes
> MiddleInitial =3D 1 byte
> SSN =3D 11 bytes
> OtherColumns =3D 258 bytes
>
> Total Bytes =3D 333 bytes
>
> Can someone explain to me the 67 bytes (400 - 333 =3D 67) I am unable to
> account for?
>
> --
> Message posted via SQLMonster.comhttp://www.sqlmonster.com/Uwe/Forums.asp=
x/sql-server/200912/1


Well first off the NCHAR are unicode, so thay actually take up 2 bytes
per carachter.  that accounts for 60..  now where is the other 7????


0
David
12/4/2009 7:51:31 PM
> Can someone explain to me the 67 bytes (400 - 333 = 67) I am unable to
> account for?

nchar is 2 bytes / char.

4 + 60 + 58 + 2 + 11 + 258 + 3 (null bitmap) + 4 (overhead) 


0
Scott
12/4/2009 8:00:19 PM
On Dec 4, 2:51=A0pm, David Hay <david....@gmail.com> wrote:
> On Dec 4, 1:50=A0pm, "cbrichards via SQLMonster.com" <u3288@uwe> wrote:
>
>
>
> > I am reading a book titled "Microsoft SQL Server 2008 Internals". In Ch=
apter
> > 6, which was authored by Kalen Delaney, Kimberly Tripp, and Paul Randal=
l, the
> > following table and Clustered Index is created:
>
> > CREATE TABLE Employee(
> > =A0 EmployeeID INT NOT NULL IDENTITY,
> > =A0 LastName NCHAR(30) NOT NULL,
> > =A0 FirstName NCHAR(29) NOT NULL,
> > =A0 MiddleInitial NCHAR(1) NULL,
> > =A0 SSN CHAR(11) NOT NULL,
> > =A0 OtherColumns CHAR(258) NOT NULL DEFAULT 'Junk')
> > GO
>
> > ALTER TABLE Employee
> > ADD CONSTRAINT EmployeePK
> > PRIMARY KEY CLUSTERED (EmployeeID)
> > GO
>
> > It states on page 321 of the aforementioned book, that "the data rows o=
f the
> > Employee table are exactly 400 bytes".
>
> > I am only computing 333 bytes, as per the following:
> > EmployeeID =3D 4 bytes
> > LastName =3D 30 bytes
> > FirstName =3D 29 bytes
> > MiddleInitial =3D 1 byte
> > SSN =3D 11 bytes
> > OtherColumns =3D 258 bytes
>
> > Total Bytes =3D 333 bytes
>
> > Can someone explain to me the 67 bytes (400 - 333 =3D 67) I am unable t=
o
> > account for?
>
> > --
> > Message posted via SQLMonster.comhttp://www.sqlmonster.com/Uwe/Forums.a=
spx/sql-server/200912/1
>
> Well first off the NCHAR are unicode, so thay actually take up 2 bytes
> per carachter. =A0that accounts for 60.. =A0now where is the other 7????

look on page 319.  not sure but what I get is the extra 7 are taken up
by the following:

status bits a - TagA byte
2 bytes to hold the number of columns
2 bytes to hold the nullbits, 1 bit for each column????
2 bytes to hold the number of var length columns

I think this is it..
0
David
12/4/2009 8:01:23 PM
Well...  LastName, FirstName, MiddleInitial are NVARCHAR, so they add up to 
120 bytes, not 60 bytes, so that brings it up to 393.

Kalen Delaney's "Inside Microsoft SQL Server 2005: The Storage Engine" book, 
page 216, I understand that the other 7 bytes are controlling information, 
therefore:  Status Bits A (1 byte), Status Bits B (1 byte), Length of fixed 
length data (2 bytes), Number of columns (2 bytes), and Null bitmap (1 
byte).

(If any columns were varying length, Status Bits A would say so and then 
there would be more controlling information and no exact data row length 
could be stated since (of course) it is varying.)

This information in also probably in the book you are reading, but I don't 
have a copy in order to point out the references.  But, from the authors' 
names alone I would recommend it.

RLF

"cbrichards via SQLMonster.com" <u3288@uwe> wrote in message 
news:a018cc86bce5e@uwe...
>I am reading a book titled "Microsoft SQL Server 2008 Internals". In 
>Chapter
> 6, which was authored by Kalen Delaney, Kimberly Tripp, and Paul Randall, 
> the
> following table and Clustered Index is created:
>
> CREATE TABLE Employee(
>  EmployeeID INT NOT NULL IDENTITY,
>  LastName NCHAR(30) NOT NULL,
>  FirstName NCHAR(29) NOT NULL,
>  MiddleInitial NCHAR(1) NULL,
>  SSN CHAR(11) NOT NULL,
>  OtherColumns CHAR(258) NOT NULL DEFAULT 'Junk')
> GO
>
> ALTER TABLE Employee
> ADD CONSTRAINT EmployeePK
> PRIMARY KEY CLUSTERED (EmployeeID)
> GO
>
> It states on page 321 of the aforementioned book, that "the data rows of 
> the
> Employee table are exactly 400 bytes".
>
> I am only computing 333 bytes, as per the following:
> EmployeeID = 4 bytes
> LastName = 30 bytes
> FirstName = 29 bytes
> MiddleInitial = 1 byte
> SSN = 11 bytes
> OtherColumns = 258 bytes
>
> Total Bytes = 333 bytes
>
> Can someone explain to me the 67 bytes (400 - 333 = 67) I am unable to
> account for?
>
> -- 
> Message posted via SQLMonster.com
> http://www.sqlmonster.com/Uwe/Forums.aspx/sql-server/200912/1
> 

0
Russell
12/4/2009 8:02:48 PM
Looks right to me!

-- 
HTH
Kalen
----------------------------------------
Kalen Delaney
SQL Server MVP
www.SQLServerInternals.com

"David Hay" <david.hay@gmail.com> wrote in message 
news:1fb2614d-6e49-433e-b9cd-2414b650de84@r5g2000yqb.googlegroups.com...
> On Dec 4, 2:51 pm, David Hay <david....@gmail.com> wrote:
>> On Dec 4, 1:50 pm, "cbrichards via SQLMonster.com" <u3288@uwe> wrote:
>>
>>
>>
>> > I am reading a book titled "Microsoft SQL Server 2008 Internals". In 
>> > Chapter
>> > 6, which was authored by Kalen Delaney, Kimberly Tripp, and Paul 
>> > Randall, the
>> > following table and Clustered Index is created:
>>
>> > CREATE TABLE Employee(
>> >   EmployeeID INT NOT NULL IDENTITY,
>> >   LastName NCHAR(30) NOT NULL,
>> >   FirstName NCHAR(29) NOT NULL,
>> >   MiddleInitial NCHAR(1) NULL,
>> >   SSN CHAR(11) NOT NULL,
>> >   OtherColumns CHAR(258) NOT NULL DEFAULT 'Junk')
>> > GO
>>
>> > ALTER TABLE Employee
>> > ADD CONSTRAINT EmployeePK
>> > PRIMARY KEY CLUSTERED (EmployeeID)
>> > GO
>>
>> > It states on page 321 of the aforementioned book, that "the data rows 
>> > of the
>> > Employee table are exactly 400 bytes".
>>
>> > I am only computing 333 bytes, as per the following:
>> > EmployeeID = 4 bytes
>> > LastName = 30 bytes
>> > FirstName = 29 bytes
>> > MiddleInitial = 1 byte
>> > SSN = 11 bytes
>> > OtherColumns = 258 bytes
>>
>> > Total Bytes = 333 bytes
>>
>> > Can someone explain to me the 67 bytes (400 - 333 = 67) I am unable to
>> > account for?
>>
>> > --
>> > Message posted via 
>> > SQLMonster.comhttp://www.sqlmonster.com/Uwe/Forums.aspx/sql-server/200912/1
>>
>> Well first off the NCHAR are unicode, so thay actually take up 2 bytes
>> per carachter.  that accounts for 60..  now where is the other 7????
>
> look on page 319.  not sure but what I get is the extra 7 are taken up
> by the following:
>
> status bits a - TagA byte
> 2 bytes to hold the number of columns
> 2 bytes to hold the nullbits, 1 bit for each column????
> 2 bytes to hold the number of var length columns
>
> I think this is it.. 

0
Kalen
12/4/2009 8:28:30 PM
Thanks, Russ!

-- 
HTH
Kalen
----------------------------------------
Kalen Delaney
SQL Server MVP
www.SQLServerInternals.com

"Russell Fields" <russellfields@nomail.com> wrote in message 
news:#dlXCzRdKHA.6048@TK2MSFTNGP04.phx.gbl...
> Well...  LastName, FirstName, MiddleInitial are NVARCHAR, so they add up 
> to 120 bytes, not 60 bytes, so that brings it up to 393.
>
> Kalen Delaney's "Inside Microsoft SQL Server 2005: The Storage Engine" 
> book, page 216, I understand that the other 7 bytes are controlling 
> information, therefore:  Status Bits A (1 byte), Status Bits B (1 byte), 
> Length of fixed length data (2 bytes), Number of columns (2 bytes), and 
> Null bitmap (1 byte).
>
> (If any columns were varying length, Status Bits A would say so and then 
> there would be more controlling information and no exact data row length 
> could be stated since (of course) it is varying.)
>
> This information in also probably in the book you are reading, but I don't 
> have a copy in order to point out the references.  But, from the authors' 
> names alone I would recommend it.
>
> RLF
>
> "cbrichards via SQLMonster.com" <u3288@uwe> wrote in message 
> news:a018cc86bce5e@uwe...
>>I am reading a book titled "Microsoft SQL Server 2008 Internals". In 
>>Chapter
>> 6, which was authored by Kalen Delaney, Kimberly Tripp, and Paul Randall, 
>> the
>> following table and Clustered Index is created:
>>
>> CREATE TABLE Employee(
>>  EmployeeID INT NOT NULL IDENTITY,
>>  LastName NCHAR(30) NOT NULL,
>>  FirstName NCHAR(29) NOT NULL,
>>  MiddleInitial NCHAR(1) NULL,
>>  SSN CHAR(11) NOT NULL,
>>  OtherColumns CHAR(258) NOT NULL DEFAULT 'Junk')
>> GO
>>
>> ALTER TABLE Employee
>> ADD CONSTRAINT EmployeePK
>> PRIMARY KEY CLUSTERED (EmployeeID)
>> GO
>>
>> It states on page 321 of the aforementioned book, that "the data rows of 
>> the
>> Employee table are exactly 400 bytes".
>>
>> I am only computing 333 bytes, as per the following:
>> EmployeeID = 4 bytes
>> LastName = 30 bytes
>> FirstName = 29 bytes
>> MiddleInitial = 1 byte
>> SSN = 11 bytes
>> OtherColumns = 258 bytes
>>
>> Total Bytes = 333 bytes
>>
>> Can someone explain to me the 67 bytes (400 - 333 = 67) I am unable to
>> account for?
>>
>> -- 
>> Message posted via SQLMonster.com
>> http://www.sqlmonster.com/Uwe/Forums.aspx/sql-server/200912/1
>>
> 
0
Kalen
12/4/2009 8:29:03 PM
Does the 3 bytes for the null bitmap equate to the number of columns (2 bytes)
and the Null bitmap (1 byte)? I ask this because page 319 states "Data rows
have an Ncol field and null bitmap whether or not any columns allow NULL". So,
I am assuming, that accounts for 3 of the missing 7 bytes.

Where do you get the 4 byte overhead?

Scott Morris wrote:
>> Can someone explain to me the 67 bytes (400 - 333 = 67) I am unable to
>> account for?
>
>nchar is 2 bytes / char.
>
>4 + 60 + 58 + 2 + 11 + 258 + 3 (null bitmap) + 4 (overhead)

-- 
Message posted via http://www.sqlmonster.com

0
cbrichards
12/4/2009 8:32:18 PM
Hi cbrichards

Check Russell's reply... he has them all listed correctly. It is also in the 
book, if you read all the details in Chapter 5.

-- 
HTH
Kalen
----------------------------------------
Kalen Delaney
SQL Server MVP
www.SQLServerInternals.com

"cbrichards via SQLMonster.com" <u3288@uwe> wrote in message 
news:a019b142729dc@uwe...
> Does the 3 bytes for the null bitmap equate to the number of columns (2 
> bytes)
> and the Null bitmap (1 byte)? I ask this because page 319 states "Data 
> rows
> have an Ncol field and null bitmap whether or not any columns allow NULL". 
> So,
> I am assuming, that accounts for 3 of the missing 7 bytes.
>
> Where do you get the 4 byte overhead?
>
> Scott Morris wrote:
>>> Can someone explain to me the 67 bytes (400 - 333 = 67) I am unable to
>>> account for?
>>
>>nchar is 2 bytes / char.
>>
>>4 + 60 + 58 + 2 + 11 + 258 + 3 (null bitmap) + 4 (overhead)
>
> -- 
> Message posted via http://www.sqlmonster.com
> 
0
Kalen
12/4/2009 8:54:38 PM
Reply:

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