Compare between byte

[chris chan]

Is there any way to compare two byte by XOR in a fast way
and return the number of bits is equal .

e.g.
b1 = 11100011
b2 = 01010010

after XOR compare will return 4 bit is equal


--

http://bbs.find-tutors.com/viewforum.php?f=30

[chris chan]

Re: Compare between byte

Sorry , It should be AND operation

--
BT.NET ¤U¤@¥N¬J BT client

http://bbs.find-tutors.com/viewforum.php?f=30

"chris chan" <50324440@stude nt.cityu.edu.hk > ¼¶¼g©ó¶l¥ó·s»D
:egC9SSQbDHA.42 4@TK2MSFTNGP10. phx.gbl...[color=blue]
> Is there any way to compare two byte by XOR in a fast way
> and return the number of bits is equal .
>
> e.g.
> b1 = 11100011
> b2 = 01010010
>
> after XOR compare will return 4 bit is equal
>
>
> --
> http://bbs.find-tutors.com/viewforum.php?f=30
>
>[/color]

[Chris Taylor]

Re: Compare between byte

Sorry,

for ( byte i = 0; i < 7; ++i )
should be
for ( byte i = 0; i < 8; ++i )

Regards

Chris Taylor

"Chris Taylor" <chris_taylor_z a@hotmail.com> wrote in message
news:esENqWSbDH A.1272@TK2MSFTN GP12.phx.gbl...[color=blue]
> Hi,
>
> I think the following function should help you achieve what you want. It
> will count the number of bits that any integer, type up to a int for this
> example, that they have in common.
>
> byte countCommonBits ( byte b1, byte b2 )
> {
> byte nCount = 0;
>
> for ( byte i = 0; i < 7; ++i )
> {
> bool r1 = ( b1 & 1 ) == 1;
> bool r2 = ( b2 & 1 ) == 1;
> if ( r1 == r2 )
> ++nCount;
> b1 >>= 1;
> b2 >>= 1;
> }
>
> return nCount;
> }
>
>
> It works by check bit 0 of both byte, if they are the same then nCount is
> incremented. Then the bit 1 is shifted into position 0 and the test is
> repeated. This is done for all 8 bits of the byte datatype.
>
> Hope this helps
>
> Chris Taylor
> "chris chan" <50324440@stude nt.cityu.edu.hk > wrote in message
> news:egC9SSQbDH A.424@TK2MSFTNG P10.phx.gbl...[color=green]
> > Is there any way to compare two byte by XOR in a fast way
> > and return the number of bits is equal .
> >
> > e.g.
> > b1 = 11100011
> > b2 = 01010010
> >
> > after XOR compare will return 4 bit is equal
> >
> >
> > --
> > http://bbs.find-tutors.com/viewforum.php?f=30
> >
> >[/color]
>
>[/color]

[chris chan]

Re: Compare between byte

thx for your reply
i will have a try ^_^

--


"Chris Taylor" <chris_taylor_z a@hotmail.com> ¼¶¼g©ó¶l¥ó·s»D
:OmSblbSbDHA.28 20@tk2msftngp13 .phx.gbl...[color=blue]
> Sorry,
>
> for ( byte i = 0; i < 7; ++i )
> should be
> for ( byte i = 0; i < 8; ++i )
>
> Regards
>
> Chris Taylor
>
> "Chris Taylor" <chris_taylor_z a@hotmail.com> wrote in message
> news:esENqWSbDH A.1272@TK2MSFTN GP12.phx.gbl...[color=green]
> > Hi,
> >
> > I think the following function should help you achieve what you want. It
> > will count the number of bits that any integer, type up to a int for[/color][/color]
this[color=blue][color=green]
> > example, that they have in common.
> >
> > byte countCommonBits ( byte b1, byte b2 )
> > {
> > byte nCount = 0;
> >
> > for ( byte i = 0; i < 7; ++i )
> > {
> > bool r1 = ( b1 & 1 ) == 1;
> > bool r2 = ( b2 & 1 ) == 1;
> > if ( r1 == r2 )
> > ++nCount;
> > b1 >>= 1;
> > b2 >>= 1;
> > }
> >
> > return nCount;
> > }
> >
> >
> > It works by check bit 0 of both byte, if they are the same then nCount[/color][/color]
is[color=blue][color=green]
> > incremented. Then the bit 1 is shifted into position 0 and the test is
> > repeated. This is done for all 8 bits of the byte datatype.
> >
> > Hope this helps
> >
> > Chris Taylor
> > "chris chan" <50324440@stude nt.cityu.edu.hk > wrote in message
> > news:egC9SSQbDH A.424@TK2MSFTNG P10.phx.gbl...[color=darkred]
> > > Is there any way to compare two byte by XOR in a fast way
> > > and return the number of bits is equal .
> > >
> > > e.g.
> > > b1 = 11100011
> > > b2 = 01010010
> > >
> > > after XOR compare will return 4 bit is equal
> > >
> > >
> > > --
> > > http://bbs.find-tutors.com/viewforum.php?f=30
> > >
> > >[/color]
> >
> >[/color]
>
>[/color]

[Mattias Sjögren]

Re: Compare between byte


If performance is important, it would probably be faster to us a
static lookup table instead of calculating the number of bits on each
call.

static int[] OnesInByte = new int[] {
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8,
};


int countCommonBits ( byte b1, byte b2 )
{
return OnesInByte[b1 & b2];
}



Mattias

--
Mattias Sjögren [MVP] mattias @ mvps.org

http://www.msjogren.net/dotnet/

Please reply only to the newsgroup.