Re: Signed multiplication

From: Ullrich von Bassewitz <>
Date: Thu, 5 Nov 2009 19:11:18 +0100
Message-ID: <>
On Thu, Nov 05, 2009 at 09:38:54AM -0800, Nate Lawson wrote:
> By "signed" do you mean simply that "result is negative if sign bits are
> different, else positive"? Or are you trying to account for overflow
> somehow?

I want the operands and the result to be treated as signed, twos complement
integers. The operands should be 16 bit (-32768 to 32767), the result should
be 32 bit wide (-2147483648 to 2147483647). A sample operation would be

        -30000 * 30000 = -900000000

For a 16x16=>16 multiplication, the standard shift-and-add algorithm works for
both, signed and unsigned integers. 

        $FFFE * $0002 = $1FFFC

if you take just the lower 16 bits of the result, it is correct for a signed

        $FFFE * $0002 = $FFFC (high 16 bits dropped)
           -2 *     2 =    -4

What I need is a multiplication that gets a correct value for the high 16 bit
(the result should be $FFFFFFFC instead of $1FFFC). Currently, I'm using an
unsigned 16x16=>32 multiplication with the absolute values of the operands,
and adjust the sign of the result. The question is, if there is a faster
method than this.



Ullrich von Bassewitz                        

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Received on 2009-11-05 19:00:03

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