Martin Kirk (mlk@asu.edu) wrote:

>Here is my situation.  The input data is 8-bit, two's complement
>data.  I need to multiply two inputs together then multiply the result
>by 8 and still get the result to fit in 16 bits.
>
>As I see it the max 8-bit value can be FFh (-1).  In order to
>make the multiply work for the 16-bit math I need to set all upper 8
>bits (I think) so that my multiplier data is FFFFh.  This will cause an
>overflow if I have two such values to multiply so I was going to limit
>one of my inputs to 5 bits (ie, max value = 1Fh).  But if I do this how
>do I use this 5-bit representation of -1 in the 16-bit adder???

Martin:

The two's-complement representation was designed to solve exactly your
problem.  If you're using narrow numbers (with a small number of
digits) and want to translate them to wider numbers (with more digits),
all you have to do is "sign-extend" your numbers.  That is, replicate
the sign bit (the narrow representation's leftmost bit) to the left,
filling all the "extra" bits in the wider representation.

In your case, you want to extend 5-bit numbers in the range [-16 - 15]
to 16-bit numbers.  Here... Take a look at the 5-bit and 16-bit
representations of some numbers in that range:

        Decimal         5-Bit Binary            16-Bit Binary
        -------         ------------            -------------
        0               00000                   00000000 00000000
        1               00001                   00000000 00000001
        10              01010                   00000000 00001010
        15              01111                   00000000 00001111
        -1              11111                   11111111 11111111
        -10             10110                   11111111 11110110
        -15             10000                   11111111 11110000

As you can see, bit-positions 5-15 in the 16-bit numbers contain the
same value (0 for positive numbers, 1 for negative) that's in bit 4.

[If you're REAL unfamiliar with binary notation, I guess I should tell
you that the bits are numbered from right to left, starting with bit
number 0.  If you're not, I apologize for talking down to you.]

If you were writing your own math routines or were more comfortable with
binary math, I'd recommend that you write your multiply routine to use
the 5-bit value directly; the resulting code would be shorter and would
execute more quickly.  However, since you're using pre-packaged
routines, just extend the sign bit into the upper bits before you do the
multiplication, and you'll be fine.

-Andy


--
Andrew Warren - fastfwd@ix.netcom.com
Fast Forward Engineering, Vista, California