[Tutor] stumped again adding bytes

[shawn bright]

Thanks for your help, Luke.
i am trying to get a grasp on how all this works, which is the msb, lsb, etc..

if i use i bitmask of 240 it will mask the most significant 4 bits
so that only the most significant 4 bits remain..
like 53 & 240 = 48 ( because only the 32 and 16 are set)
and if i use 15 it should mask the least significant bits, right ?

now if i am using the first half of the 2nd byte, and making the first
byte the most significant.
i would shift the first byte to the Left by 4
byte1 << 4  then add ( byte2 & 240 )

so for the next number that i want that will consist of the 2nd half
of the 2nd byte as the most significant , i would take the second byte
with my bitmask then shift it right by 4 then add the third byte to
that .... am i getting this right
like x = ((byte2 & 240) << 4) + byte 3

i think ?

shawn


On 2/23/07, Luke Paireepinart <rabidpoobear at gmail.com> wrote:
> shawn bright wrote:
> > ok, i am good with what you have explained here,
> > now i am on a similar problem.
> >
> > the data i need is in a condensed format. in other words, they are
> > sending 2 values in three bytes.
> >
> > so if i have 3 values say a = 53, b = 13, and c = 31
> >
> > so value 1 is the first byte ( a ) and the first 4 bits of the second
> > byte (b)
> > value 2 is the last 4 bits of byte (b) and byte (c)
> Please try not to interchange your terms, it makes your questions confusing.
> 'so if i have 3 values say a = 53, b = 13, and c = 31'
> Should be 'so if I have 3 bytes'
> because immediately after, you talk about your extracted data as 'values.'
> It's not too big of a deal, but it was confusing at first.
> >
> > so i believe i do shifting here. as in i do a
> > (a << 4) * 32 + b
> You need to read an article on bit manipulation.
> I'll give you a summary of what you're trying to do.
> given the byte
> x = 1 1 0 0 0 1 1 0
> If you want to extract just the four most significant bits, [1 1 0 0],
> you shift the number TO THE RIGHT by 4.
> x >> 4
> will result in 4 shifts like so:
> 1 1 0 0 0 1 1 0 -> 1 1 0 0 0 1 1
> 1 1 0 0 0 1 1 -> 1 1 0 0 0 1
> 1 1 0 0 0 1 -> 1 1 0 0 0
> 1 1 0 0 0 -> 1 1 0 0
> Which results in the number 0x0C or 12.
> Now, since these are the MSB, you could shift back 4 places
> x << 4
> 1 1 0 0 -> 1 1 0 0 0
> 1 1 0 0 0 -> 1 1 0 0 0 0
> 1 1 0 0 0 0 -> 1 1 0 0 0 0 0
> 1 1 0 0 0 0 0 -> 1 1 0 0 0 0 0 0
>
> but I don't know if you want to do that or not.
> and if you wanted to do that, you could just use a bitmask from the
> beginning.  see the next part.
>
> Now as for the least significant bits:
> recall that
>
> 1 1 0 0 0 1 0 0 & 0 0 0 0 1 1 1 1
> will yeild
> 0 0 0 0 + whatever the last 4 bits are in the first item.
>
> Hope that helps.
> It's better if you take the time to really understand what is happening
> in all these cases, so you won't have to experiment with trial and error.
> -Luke
>