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An introduction to bitwise operators

By , 8 May 2002
 

Introduction

I have noticed that some people seem to have problems with bitwise operators, so I decided to write this brief tutorial on how to use them.

An Introduction to bits

Bits, what are they you may ask?

Well, simply put, bits are the individual ones and zeros that make up every thing we do with computers. All the data you use is stored in your computer using bits. A BYTE is made up of eight bits, a WORD is two BYTEs, or sixteen bits. And a DWORD is two WORDS, or thirty two bits.

 0 1 0 0 0 1 1 1 1 0 0 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 1 1 1 0 0 0
||              |               |               |              ||
|+- bit 31      |               |               |       bit 0 -+|
|               |               |               |               |
+-- BYTE 3 -----+--- BYTE 2 ----+--- BYTE 1 ----+-- BYTE 0 -----+
|                               |                               |
+----------- WORD 1 ------------+----------- WORD 0 ------------+
|                                                               |
+--------------------------- DWORD -----------------------------+

The beauty of having bitwise operators is that you can use a BYTE, WORD or DWORD as a small array or structure. Using bitwise operators you can check or set the values of individual bits or even a group of bits.

Hexadecimal numbers and how they relate to bits

When working with bits, it is kind of hard to express every number using just ones and zeros, which is known as binary notation. To get around this we use hexadecimal (base 16) numbers.

As you may or may not know, it takes four bits to cover all the numbers from zero to fifteen, which also happens to be the range of a single digit hexadecimal number. This group of four bits, or half a BYTE, is called a nibble. As there are two nibbles in a BYTE, we can use two hexadecimal digits to show the value of one BYTE.

NIBBLE   HEX VALUE
======   =========
 0000        0
 0001        1
 0010        2
 0011        3
 0100        4
 0101        5
 0110        6
 0111        7
 1000        8
 1001        9
 1010        A
 1011        B
 1100        C
 1101        D
 1110        E
 1111        F

So if we had one BYTE containing the letter 'r' (ASCII code 114) it would look like this:

0111 0010    binary
  7    2     hexadecimal

We could write it as '0x72'

Bitwise operators

There are six bitwise operators. They are:
   &   The AND operator
   |   The OR operator
   ^   The XOR operator
   ~   The Ones Complement or Inversion operator
  >>   The Right Shift operator
  <<   The Left Shift operator.

The & operator

The & (AND) operator compares two values, and returns a value that has its bits set if, and only if, the two values being compared both have their corresponding bits set. The bits are compared using the following table

   1   &   1   ==   1
   1   &   0   ==   0
   0   &   1   ==   0
   0   &   0   ==   0

An ideal use for this is to set up a mask to check the values of certain bits. Say we have a BYTE that contains some bit flags, and we want to check if bit four bit is set.

BYTE b = 50;
if ( b & 0x10 )
    cout << "Bit four is set" << endl;
else
    cout << "Bit four is clear" << endl;

This would result in the following calculation

    00110010  - b
  & 00010000  - & 0x10
  ----------
    00010000  - result

So we see that bit four is set.

The | operator

The | (OR) operator compares two values, and returns a value that has its bits set if one or the other values, or both, have their corresponding bits set. The bits are compared using the following table

   1   |   1   ==   1
   1   |   0   ==   1
   0   |   1   ==   1
   0   |   0   ==   0

An ideal use for this is to ensure that certain bits are set. Say we want to ensure that bit three of some value is set

BYTE b = 50;
BYTE c = b | 0x04;
cout << "c = " << c << endl;

This would result in the following calculation

    00110010  - b
  | 00000100  - | 0x04
  ----------
    00110110  - result

The ^ operator

The ^ (XOR) operator compares two values, and returns a value that has its bits set if one or the other value has its corresponding bits set, but not both. The bits are compared using the following table

   1   ^   1   ==   0
   1   ^   0   ==   1
   0   ^   1   ==   1
   0   ^   0   ==   0

An ideal use for this is to toggle certain bits. Say we want toggle the bits three and four

BYTE b = 50;
cout << "b = " << b << endl;
b = b ^ 0x18;
cout << "b = " << b << endl;
b = b ^ 0x18;
cout << "b = " << b << endl;

This would result in the following calculations

    00110010  - b
  ^ 00011000  - ^ 0x18
  ----------
    00101010  - result

    00101010  - b
  ^ 00011000  - ^ 0x18
  ----------
    00110010  - result

The ~ operator

The ~ (Ones Complement or inversion) operator acts only on one value and it inverts it, turning all the ones int zeros, and all the zeros into ones. An ideal use of this would be to set certain bytes to zero, and ensuring all other bytes are set to one, regardless of the size of the data. Say we want to set all the bits to one except bits zero and one

BYTE b = ~0x03;
cout << "b = " << b << endl;
WORD w = ~0x03;
cout << "w = " << w << endl;

This would result in the following calculations

    00000011  - 0x03
    11111100  - ~0x03  b

    0000000000000011  - 0x03
    1111111111111100  - ~0x03  w

Another ideal use, is to combine it with the & operator to ensure that certain bits are set to zero. Say we want to clear bit four

BYTE b = 50;
cout << "b = " << b << endl;
BYTE c = b & ~0x10;
cout << "c = " << c << endl;

This would result in the following calculations

    00110010  - b
  & 11101111  - ~0x10
  ----------
    00100010  - result

The >> and << operators

The >> (Right shift) and << (left shift) operators move the bits the number of bit positions specified. The >> operator shifts the bits from the high bit to the low bit. The << operator shifts the bits from the low bit to the high bit. One use for these operators is to align the bits for whatever reason (check out the MAKEWPARAM, HIWORD, and LOWORD macros)

BYTE b = 12;
cout << "b = " << b << endl;
BYTE c = b << 2;
cout << "c = " << c << endl;
c = b >> 2;
cout << "c = " << c << endl;

This would result in the following calculations

    00001100  - b
    00110000  - b << 2
    00000011  - b >> 2

Bit Fields

Another interesting thing that can be done using bits is to have bit fields. With bit fields you can set up minature structures within a BYTE, WORD or DWORD. Say, for example, we want to keep track of dates, but we want to use the least amount of memory as possible. We could declare our structure this way

struct date_struct {
    BYTE day   : 5,   // 1 to 31
         month : 4,   // 1 to 12
         year  : 14;  // 0 to 9999
    } date;

In this example, the day field takes up the lowest 5 bits, month the next four, and year the next 14 bits. So we can store the date structure in twenty three bits, which is contained in three BYTEs. The twenty fourth bit is ignored. If I had declared it using an integer for each field, the structure would have taken up 12 BYTEs.

|0 0 0 0 0 0 0 0|0 0 0 0 0 0 0 0|0 0 0 0 0 0 0 0|
  |                           |       |         |
  +------ year ---------------+ month +-- day --+

Now lets pick this declaration apart to see what we are doing.

First we will look at the data type we are using for the bit field structure. In this case we used a BYTE. A BYTE is 8 bits, and by using it, the compiler will allocate one BYTE for storage. If however, we use more than 8 bits in our structure, the compiler will allocate another BYTE, as many BYTEs as it takes to hold our structure. If we had used a WORD or DWORD, the compiler would have allocated a total of 32 bits to hold our structure.

Now lets look at how the various fields are declared. First we have the variable (day, month, and year), followed by a colon that separates the variable from the number of bits that it contains. Each bit field is separated by a comma, and the list is ended with a semicolon.

Now we get to the struct declaration. We put the bit fields into a struct like this so that we can use convention structure accessing notation to get at the structure members. Also, since we can not get the addresses of bit fields, we can now use the address of the structure.

date.day = 12;

dateptr = &date;
dateptr->year = 1852;

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

About the Author

PJ Arends
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GeneralRe: Greatmembermanzoor1012 Dec '08 - 4:12 
I'm programming for more than 4 months and haven't understood them fully, thanks for this article it removed few of my confusions.
 
http://www.webhostingyes.com/

Generalwe did it againmemberJeremy Falcon25 Jun '03 - 5:33 
Well, looks like we have similar articles again! Great minds think alike, eh? Smile | :)
 
Anyway, since I believe that our articles complement one another, would you want to cross-link them? Also, would you mind if I referenced your article in my future ones that will be meant to build upon my Learning Binary and Hexadecimal one?
 
BTW, if you'd like, I'd also be willing to make a downloadable PDF version for you. I have Acrobat 5 at home and just look for excuses to use it. Smile | :)
 
Jeremy Falcon
GeneralRe: we did it againmemberPJ Arends25 Jun '03 - 6:54 
Jeremy Falcon wrote:
Well, looks like we have similar articles again! Great minds think alike, eh?
 
Yeah, but mine was first this time;P
 
Jeremy Falcon wrote:
Anyway, since I believe that our articles complement one another, would you want to cross-link them? Also, would you mind if I referenced your article in my future ones that will be meant to build upon my Learning Binary and Hexadecimal one?
 
You can reference this one from your articles if you like, but I will consider your note here to be a cross reference to your article. (I'm too damn lazy to bother updating my article)
 
Jeremy Falcon wrote:
BTW, if you'd like, I'd also be willing to make a downloadable PDF version for you. I have Acrobat 5 at home and just look for excuses to use it.
 
It is nice of you offer, but until PDF versions of articles becomes the CP standard I will pass. But do feel free to make a PDF copy for your own personal use if you desireSmile | :)
 






Sonork 100.11743 Chicken Little
 
"You're obviously a superstar." - Christian Graus about me - 12 Feb '03
 
Within you lies the power for good - Use it!
GeneralNibble SwapsussBcdarus1 Jun '03 - 18:36 
Hi,
I want to know how to swap nibbles in a byte without using another variable.
 
For example,
 
X = 0xAF,
 
then swapped value should be
X = 0xFA
 
And this should be done without using any other variable.
 
Regards,
Bcdarus
GeneralRe: Nibble SwapmemberPJ Arends2 Jun '03 - 5:40 
I don't do other peoples homework, but I will give you a clue: You need to use the &, |, <<, and >> operators.
 
If you post some code you already have, I may be able to help you some moreSmile | :)
 






Sonork 100.11743 Chicken Little
 
"You're obviously a superstar." - Christian Graus about me - 12 Feb '03
 
Within you lies the power for good - Use it!
GeneralASL vs ROLmemberKochise25 Jun '03 - 5:21 
In Motorola assembly, try :
 
move.b  #$AF,D0
asl.b   #4,D0
 
and
 
move.b  #$AF,D0
rol.b   #4,D0
 
The first returns $F0 (puts four 0 bits from the right), the second returns $FA (the outputing four $F bits from the left are reinserted from the right)...
 
In C, when using <<, it's like using ASL (Arithmetic Shift Left). There is nothing like ROL (ROtate Left) in C...
 
Use also something like :
 
unsigned char nSwap;
nSwap = 0xAF;
nSwap = (nSwap << 4) || (nSwap >> 4); // 0xF0 || 0x0A
 
Kochise
 
In Cod we trust !
GeneralRe: ASL vs ROLmemberPJ Arends25 Jun '03 - 6:40 
The code I was thinking of is:
x = ((x & 0x0F) << 4) | ((x & 0xF0) >> 4);
I don't think your use of the logical-OR ( || ) operator will work, but I have not tried it;)
 






Sonork 100.11743 Chicken Little
 
"You're obviously a superstar." - Christian Graus about me - 12 Feb '03
 
Within you lies the power for good - Use it!
GeneralMistyping...memberKochise25 Jun '03 - 22:49 
Yeap, |, not || Wink | ;)
 
Just add that the '& 0x0F' and '& 0xF0' are useless if you work on a char (8 bits bounding)
 
Kochise
 
In Cod we trust !
GeneralRe: Mistyping...memberRyan Binns30 Jun '03 - 23:02 
Kochise wrote:
Just add that the '& 0x0F' and '& 0xF0' are useless if you work on a char (8 bits bounding)
 
Not quite. It depends on whether the compiler performs a logical right-shift or an arithmetic right-shift. For instance:
 
0xfa >> 4 using a logical shift gives 0x0f, as required
0xfa >> 4 using an arithmetic shift gives 0xff, certainly not wanted.
 
Logical shifts shift in zero as the top bit. Arithmetic shifts copy the old top bit as the new top bit, as can be seen in the above example.
 
There is no requirement for the compiler to choose one over the other, so leaving out the '&' bit is just asking for trouble. I believe it works with the MS compiler, but I'm not sure about any others.
 
Ryan
 
Being little and getting pushed around by big guys all my life I guess I compensate by pushing electrons and holes around. What a bully I am, but I do enjoy making subatomic particles hop at my bidding - Roger Wright (2nd April 2003, The Lounge)

Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late - John Nichol "Point Of Impact"

GeneralRe: Mistyping...memberKochise1 Jul '03 - 0:26 
Right, but the QNX GCC and the Lattice C compilers do the same, ONLY logical shift... Even in the GreenLeaf compiler there is NO arithmetic shift !
 
But true, to be sure whet is done, the best is to force the '& 0x0F' and '& 0xF0' in every case !
 
Kochise
 
In Cod we trust !
General* bookmarked *memberNitron4 Feb '03 - 4:33 
Ahh, the article that keeps on giving! A must have for every professional! Thanx!
 
Rose | [Rose]
 
- Nitron


"Those that say a task is impossible shouldn't interrupt the ones who are doing it." - Chinese Proverb
 

GeneralGood Starting pointmemberKarstenK12 Nov '02 - 1:26 
I like this article a lot. It´s a good repetition about bits and its operators.
But really interesting it became at the date-struct.
 
I think there could be a "bits revisited" article. Theres a lot of castings and macros used in the API and MFC you better should know. Blush | :O
 
I do also a lot of enums with bitfields to separater different states of operation. Cool | :cool:

 
Try it @ home. (B&B)
GeneralThanks a lotmemberpoplarc3 Nov '02 - 19:54 
It's very useful for a beginner just like me.
But, what is the Opps's meaning?Smile | :)
(Sorry for my poor English.Blush | :O )
GeneralGood articlememberNilesh K. Karkhanis10 Oct '02 - 23:45 
Keep it up Smile | :)
 
Ciao
 
- Nilesh
GeneralGsm softwarememberloni8 Sep '02 - 8:29 
Did anyone now s about gsm softwares??
Confused | :confused:
GeneralBIG Endian vs little EndianmemberKochise22 Aug '02 - 23:09 

In other words, Motorola vs Intel !
 
What the deal about these Endians ? I not said Indians Wink | ;) It's the way data is stored in MEMORY. Well, numbers such as a long/DWORD (32 bits) 0x12345678 are ALWAYS stored in BIG ENDIAN into the microprocessor registers, but in a different way into the memory according to the microprocessor firm.
 
Data are sized differently :
Byte : One byte, noted byte 0
Word : Two bytes, from byte 0 to 1
Long : Four bytes, from byte 0 to 3
 
Every byte is formed with 8 bits, from bit 7 to 0 (left to right).
Every word is formed with 16 bits, from bit 15 to 0 (left to right)..
Every long is formed with 32 bits, from bit 31 to 0 (left to right)..
 
Note that byte 0 is the LSB (Less Significant Byte) and byte 3 is the MSB (Most Significant Byte). So :
Byte : 3 2 1 0
Number : $12 $34 $56 $78
 
The number given is $12*256^3 + $34*256^2 + $56*256^1 + $78*256^0.
 
It works also for words, word 0 is the 'LSW' (Less Significant Word) and word 1 is the 'MSW' (Most Significant Word) :
Word : 1 0
Number : $1234 $5678
 
The number given is $1234*65536^1 + $5678*65536^0.
 
Now the way bytes are stored :
 
BIG ENDIAN : First END is BIG power, that's to say BIG byte first...
Byte : Byte 3 ($12) (just MSB)
Word : Byte 3 to 2, word 1 ($1234) (MSB to LSB)
Long : Byte 3 to 0, word 1 to 0 ($12345678) (MSB to LSB)
 
BIG ENDIAN is the way we ALL write numbers on paper, most significant/powerful digit on the left (for instance thousands, then hundreds, then units). It's also the way ALL DATAS are stored in registers, for ALL processors architechtures !
 
Into memory, you'll find at the address 'n'(+ offset) the following data :
n+0 : Byte 3 ($12) (MSB first : BIG Endian)
n+1 : Byte 2 ($34)
n+2 : Byte 1 ($56)
n+3 : Byte 0 ($78) (LSB)
 
Byte 0 location : 'n+3' (just 'n+3', byte 0 is $78)
Word 0 location : 'n+2' ('n+2' to 'n+3', byte 1 to byte 0, word 0 is $5678)
Long 0 location : 'n+0' ('n+0' to 'n+3', byte 3 to byte 0, long 0 is $12345678)
 
It's really useful when you work in embedded systems and you debug memory mappings. What you have in registers is what you get in memory. But fetching the Low Word of the data needs to fetch it at 'n+2'. So the fetching is a bit slower due to the additive memory address shifting to fetch the 'casted' (according to the assembly opcode operand size) data.
 
Into register : $12345678
At address 'n+0' : $12 $34 $56 $78 (MSB to LSB per byte, byte 3 to 0)
At address 'n+0' : $1234 $5678 (MSB to LSB per word, word 1 to 0)
 
BIG ENDIAN numbers are also widely used into TCP/IP protocols ! The most common BIG Endian architechtures are Motorola 6800 family, 68000 family and ColdFire family. The PowerPC is a little Endian one (derivated from the IBM Power processor family).
 
LITTLE ENDIAN : First END is LITTLE power, that's to say little byte first...
Byte : Byte 0 ($78) (just LSB)
Word : Byte 0 to 1, word 0 ($7856) (LSB to MSB)
Long : Byte 0 to 3, word 0 to 1 ($78563412) (LSB to MSB)
 
LITTLE ENDIAN processors are hardwired to convert LITTLE ENDIAN from the memory to BIG ENDIAN into registers. So it takes no more time to load/store data on LITTLE ENDIAN processors, don't worry...
 
Into memory, you'll find at the address 'n' (+ offset) the following data :
n+0 : Byte 0 ($78) (LSB first : little Endian)
n+1 : Byte 1 ($56)
n+2 : Byte 2 ($34)
n+3 : Byte 3 ($12) (MSB)
 
Byte 0 location : 'n+0' (just 'n+0', byte 0 is $78)
Word 0 location : 'n+0' ('n+0' to 'n+1', byte 0 to 1, word 0 is $5678 in LITTLE ENDIAN)
Long 0 location : 'n+0' ('n+0' to 'n+3', byte 0 to 3, long 0 is $12345678 in LITTLE ENDIAN)
 
It is useful when you want to get the low part of the data, for instance the LSB or the LSW. You just have to fetch the 'casted' (according to the assembly opcode operand size) data at the same address 'n+0' ! But debuging memory is then absolutely weird...
 
Into register : $12345678
At address 'n+0' : $78 $56 $34 $12 (LSB to MSB per byte in LITTLE ENDIAN)
At address 'n+0' : $7856 $3412 (LSB to MSB per word in LITTLE ENDIAN)
 
Please note that some few processors uses bitwise LITTLE ENDIAN storage, that's to say a 32 bits long from bit 31 to 0 is stored in memory from bit 0 to 31 !
 
Into register : $12345678 (%10010001101000101011001111000)
At address 'n+0' : $01CD4589 (%0001110011010100010110001001) (GOD, these fools !)
 
Kochise
 
PS : The way numbers format is noted...
 
Firm : Motorola Toshiba Intel
Decimal : 123 123 123 (default)
Octal : @173 o173 173o
Hexa : $7B h7B 7Bh
Binary : %1111011 b1111011 1111011b
 
My favorite way is the Motorola's since it is easier to first know in which format is written the number, then reading the number. For instance, when I read $1234, I immediatly know that it means 4660 in decimal format. Harder to first read 1234, then discover the number was given in hexadecimal (Intel). Toshiba's way is good also, but I sometimes find hard to read o01101 or h481 ! Characters @, $ and % cannot be missed !

The useful links :
 
http://facstaff.bloomu.edu/bobmon/readings/ien137.Cohen-Holy_Wars.html[^] (THE MUST TO READ)
 
http://www.webopedia.com/TERM/b/big_endian.html[^]
http://www.cs.umass.edu/~verts/cs32/endian.html[^]
http://www.cscc.de/download/java/big-endian-versus-little-endian.txt[^]
http://www.rfc-editor.org/rfc/rfc1071.txt[^] (essential read)
http://ourworld.compuserve.com/homepages/rortiz/LEndians.htm[^]
http://www.ddj.com/ftp/1999/1999_03/antlr.txt[^] (Example 6)
http://www.geocities.com/fontboard/cjk/unicode.html[^]
GeneralHow to know?memberrichard sancenot8 Sep '05 - 23:00 
Well, first thanks for this article.Rose | [Rose]
 
Here is my issue :
 
I receive data from RS232 serial port, i know that those data are represented in little-endian format.
 
Then, I directly copy those binaries data into the mermory to end up with a filled c++ structure.
 
My fear is that my processor interprets a DWORD differently depending on the way data is stored in the memory.
 
How am I supposed to know if a processor uses little/big endian Confused | :confused: ?
Maybe it's a false problem, but let me know Smile | :) .
 
Thanks.
 
Tout programme dont la fiabilité dépend de l'homme n'est pas fiable
GeneralA minor correction...memberGurj13 Jun '02 - 12:54 
Great article...
 
However, a minor correction is needed: (so as not to confuse a beginner Smile | :) )
-------------------------------------
In the | operator usage example, the mask for setting the 3rd bit should be 0x08 [assuming bits
are counted starting from 0 (i.e., zero) as illustrated in the rest of this article]
 
i.e.,
An ideal use for this is to ensure that certain bits are set. Say we want to ensure that bit three of some value is set
 
BYTE b = 50;
BYTE c = b | 0x08;
cout << "c = " << c << endl;
This would result in the following calculation
 
  00110010 - b
| 00001000 - | 0x08
----------
  00111010 - result

GeneralRe: A minor correction...memberPJ Arends13 Jun '02 - 14:18 
Opps.Blush | :O
 
I think it's time I fired my proof reader.
 
---
CPUA 0x5041
 
Sonork 100.11743 Chicken Little
 
Within you lies the power for good - Use it!
GeneralThanks......memberMazdak23 May '02 - 21:52 
I have a DWORD type and I wanna check if its bits set(e.g bit number 10).
 
DWORD dword = somevalue;
if(dword & here I should give 10 in HEX or another way????(DWORD is 32 bit))
  do something
 
Am I in correct way?
 
Thanks
 
<html>Mazy</html>
 
"The more I search, the more my need
For you,
The more I bless, the more I bleed
For you."The Outlaw Torn-Metallica


GeneralRe: Thanks......memberPJ Arends24 May '02 - 6:23 
If you want to check bit 10 you need to do this:
 
1.) Find bit #10
                         This is bit #10 --+
                                           |
 0 1 0 0 0 1 1 1 1 0 0 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 1 1 1 0 0 0
||              |               |               |              ||
|+- bit 31      |               |               |       bit 0 -+|
|               |               |               |               |
+-- BYTE 3 -----+--- BYTE 2 ----+--- BYTE 1 ----+-- BYTE 0 -----+
|                               |                               |
+----------- WORD 1 ------------+----------- WORD 0 ------------+
|                                                               |
+--------------------------- DWORD -----------------------------+
2.) Convert that to a hex number with only bit #10 set
0x00000400
 
3.) Use the & (AND) operator to see if the bit is set
if (dword & 0x00000400)
{
    DoSomething()
}
I hope this makes things a little clearer for you.Big Grin | :-D
 
---
CPUA 0x5041
 
Sonork 100.11743 Chicken Little
 
Orgasms are universal!! OMG | :OMG: -- Mike Mullikin in The Lounge 21:27 15 May '02
GeneralRe: Thanks......memberMazdak24 May '02 - 8:42 
PJ Arends wrote:
I hope this makes things a little clearer for you
 
Yes,thank you. Smile | :)
 
<html>Mazy</html>
 
"The more I search, the more my need
For you,
The more I bless, the more I bleed
For you."The Outlaw Torn-Metallica


GeneralRe: Thanks......memberTim Smith24 May '02 - 8:52 
You can also do this...
 
if (dwValueIAmChecking & (1 << 10))
{
}
 
It looks strange if you have never seen it before, but it does help to reduce the conversion bugs when converting from a bit number of a hex constant.
 
Tim Smith
 
I know what you're thinking punk, you're thinking did he spell check this document? Well, to tell you the truth I kinda forgot myself in all this excitement. But being this here's CodeProject, the most powerful forums in the world and would blow your head clean off, you've got to ask yourself one question, Do I feel lucky? Well do ya punk?
GeneralRe: Thanks......memberMazdak24 May '02 - 19:07 
Thank you Tim. Smile | :)
GeneralGreat but....memberjohny quest19 May '02 - 18:16 
Gr88888888888888 article.I think there should be more explanation about shift operators. Can u tell me how to reverse the bytes in DWORD i.e. interchanging the position s of bytes in DWORD, last bytes come in place of first byte and so on(completely reverse the DWORD bytes).Smile | :)

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