"gene kelley" <o***@by.me> schreef in bericht
news:c1jhl2hqt03mg1q120o8uk9dv8de6u575n@4ax.com...
>I have an application where I need to read some header information found in
>3
> different types of file types.  Two of the file types were fairly straight
> forward as
> the items to read  in the header are at least 8 bits (one byte), so, I'm
> able to step
> through a file stream with a binary reader and retrive the data.  The last
> file type,
> however, has me stumped at the moment.  The header spec specifies the item
> lengths in
> bits.  Most of the item lengths are 8 bit multiples which is not a
> problem.  There
> are three items, however, that are listed with less than 8 bit lengths -
> one as 1
> bit, one as 7 bit, and another at 3 bits.
>
> In VB, how do you read bit values?
>
> Gene
>

On 13 Nov 2006 22:20:58 -0800, "Branco Medeiros" <branco.medei***@gmail.com> wrote:

>gene kelley wrote:
>> I have an application where I need to read some header information found in 3
>> different types of file types.  Two of the file types were fairly straight forward as
>> the items to read  in the header are at least 8 bits (one byte), so, I'm able to step
>> through a file stream with a binary reader and retrive the data.  The last file type,
>> however, has me stumped at the moment.  The header spec specifies the item lengths in
>> bits.  Most of the item lengths are 8 bit multiples which is not a problem.  There
>> are three items, however, that are listed with less than 8 bit lengths - one as 1
>> bit, one as 7 bit, and another at 3 bits.
>>
>> In VB, how do you read bit values?
>
>Usually it consists on using the And operator and the Shift operators
>(<< and >>). But, in the end, it all will depend on how the sequence of
>elements is layed out in the byte stream.
>
>Just to give you a starting point, consider that to read an arbitrary
>amount of bits from a byte you need first to isolate the relevant bits
>and align them to the lower boundary of the byte.
>
>Suppose for instance you have a 3-bit value. The easier situation is
>when the bits are already "right-aligned" in the byte:
>
>   aaaaabbb
>
>(where 'a' represents "don't care" bits, and 'b' represents the bits
>you want).
>
>In this case you need only to mask the relevant bits with an And-mask,
>that is, a sequence of n 1 bits, where n is the size of your data -- 3,
>in this case:
>
>           aaaaabbb And 00000111 = 00000bbb
>
>How to find the mask, you ask? Easy:
>
>   Dim Mask As Byte = (1 << Size) - 1
>
>The first problem begins when the bits are not 'right-aligned' in the
>byte:
>
>   aaabbbaa
>
>To resolve this you must first put the bits in the 'right' position.
>You do this by shifting the bits to right n positions, where n is the
>leftmost position of the bit pattern plus one, minus the size of the
>pattern:
>
>  Dim Shift As Byte = LeftmostBit + 1 - Size
>  Value = Value >> Shift
>
>Notice, however, that the bit positions are numbered like this:
>
>   76543210
>
>Therefore, in the previous example, the leftmost bit would be 4.
>
>So, to put it all together so far you'd have something like:
>
>  <aircode>
>  Function GetBits(Value as Byte, _
>  LeftmostBit As Byte, _
>  Size as Byte) As Byte
>    Dim Shift As Byte = LeftmostBit + 1 - Size
>    Dim Mask As Byte = (1 << Size) - 1
>    Return (Value >> Shift) And Mask
>  End Function
>  </aircode>
>
>Probably, depending on your settings, VB will give you all types of
>warnings, because in the above expressions the value "1" is computed as
>an integer, and the entire expression is promoted to integer. Getting
>rid of the warnings remains as an exercise =))
>
>Finally, things become *really* complicated when you have a stream of
>values whose bits span two bytes:
>
>   aaaaaabb   baaaaaaaa
>
>To resolve this, you must treat the two bytes as a 16 bit value (a
>Short, in VB). When you do this, everything else Just Works (TM), as
>long as you treat the leftmost bit position as a number from 0 to 15
>(and not 0 to 7, as in the previous example). If we use hexadecimal to
>represent the bit positions, we'd have:
>
>  FEDCBA98 76543210
>
>Therefore, in the above example, our leftmost bit would be 9 !
>
>A function that would extract a given bit pattern from a sequence of
>two bytes could be like this:
>
>  <aircode>
>  Function GetBits(FirstByte As Byte, _
>  ScndByte As Byte, _
>  LeftmostBit As Byte, _
>  Size As Byte) As Byte
>    Dim Value As Integer = (CInt(FirstByte) << 8) Or ScndByte
>    Dim Shift As Integer = LeftmostBit + 1 - Size
>    Dim Mask As Integer = (1 << Size) - 1
>    Dim Result As Integer = (Value >> Shift) And Mask
>    Return CByte(Result And 255)
>  End Function
>  </aircode>
> 
>HTH.
>
>Regards,
>
>Branco.

"gene kelley" <o***@by.me> wrote in message
news:c1jhl2hqt03mg1q120o8uk9dv8de6u575n@4ax.com...
>I have an application where I need to read some header information found in
>3
> different types of file types.  Two of the file types were fairly straight
> forward as
> the items to read  in the header are at least 8 bits (one byte), so, I'm
> able to step
> through a file stream with a binary reader and retrive the data.  The last
> file type,
> however, has me stumped at the moment.  The header spec specifies the item
> lengths in
> bits.  Most of the item lengths are 8 bit multiples which is not a
> problem.  There
> are three items, however, that are listed with less than 8 bit lengths -
> one as 1
> bit, one as 7 bit, and another at 3 bits.
>
> In VB, how do you read bit values?
>
> Gene
>

>I have an application where I need to read some header information found in 3
>different types of file types.  Two of the file types were fairly straight forward as
>the items to read  in the header are at least 8 bits (one byte), so, I'm able to step
>through a file stream with a binary reader and retrive the data.  The last file type,
>however, has me stumped at the moment.  The header spec specifies the item lengths in
>bits.  Most of the item lengths are 8 bit multiples which is not a problem.  There
>are three items, however, that are listed with less than 8 bit lengths - one as 1
>bit, one as 7 bit, and another at 3 bits.
>
>In VB, how do you read bit values?
>
>Gene
>
> 
>

>You could just declare a byte, then assign it a byte value that
>corresponds to the bit positions, then use an AND to see what your
>header value is.  Example:
>Suppose the value read from your header is thevalue
>If you want to compare the first 3 bits
>Dim theByte as Byte = 8
>Dim theResult in Byte
>A value of 8 is like saying in binary: 00000111
>
>So, theByte = (thevalue AND theByte)
>Now work with the resultant value of theByte
>
>If you want to ignore the first bit and look at the next 7:
>theByte = 254
>That's like saying in binary 11111110
>Then use your AND operator again
>
>I think this is a lot easier than shifting bits around.
>
>By the way, you would follow the same type of thing with 16 bit and 32
>bit values, but they would be easier to work with in HEX.  You could use
>HEX with the Byte too - &h08 would still check the first 3 bits, &hfe
>would compare those 7 bits.
>
>T

>
>gene kelley wrote:
>
>>I have an application where I need to read some header information found in 3
>>different types of file types.  Two of the file types were fairly straight forward as
>>the items to read  in the header are at least 8 bits (one byte), so, I'm able to step
>>through a file stream with a binary reader and retrive the data.  The last file type,
>>however, has me stumped at the moment.  The header spec specifies the item lengths in
>>bits.  Most of the item lengths are 8 bit multiples which is not a problem.  There
>>are three items, however, that are listed with less than 8 bit lengths - one as 1
>>bit, one as 7 bit, and another at 3 bits.
>>
>>In VB, how do you read bit values?
>>
>>Gene
>>
>> 
>>