Date: Wed, 15 Nov 2000 19:24:43 -0800 (PST) From: John Polstra <jdp@polstra.com> To: net@freebsd.org Cc: louie@TransSys.COM Subject: Re: libalias: Incremental Update of Internet Checksum Message-ID: <200011160324.eAG3Ohi47268@vashon.polstra.com> In-Reply-To: <200011152043.eAFKhMG68318@whizzo.transsys.com> References: <Pine.BSF.4.21.0011130015100.50906-100000@carcassonne.scientech.com> <200011151548.eAFFmJG66031@whizzo.transsys.com> <20001115180842.A11913@sunbay.com> <200011152043.eAFKhMG68318@whizzo.transsys.com>
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In article <200011152043.eAFKhMG68318@whizzo.transsys.com>,
Louis A. Mamakos <louie@TransSys.COM> wrote:
>
> > You mentioned the books... Can you citate from one of these books that
> > states that 0xFFFF + 0x0001 in one's complement arithmetic is 0x0000
> > rather than 0xFFFF?
>
> The correct result is 0x0001, since 0xffff is -0, and -0 + 1 == 1.
> 0xFFFE + 0x0001 == 0, because 0xFFFE is -1 and -1 + 1 == 0.
Louie is correct. (And in case there's any doubt, my HP 16c confirms
it in 16-bit 1's complement mode.) To add two numbers in 1's
complement you do this:
1. Perform a normal 2's complement add, and keep track of whether
there was a carry out of the high-order bit.
2. If there was a carry, increment (2's complement) the result.
In other words:
u_int16_t
add1(u_int16_t a, u_int16_t b)
{
u_int32_t sum;
sum = (u_int32_t)a + (u_int32_t)b; /* Carry goes to bit 16 */
sum += sum >> 16; /* Add in the carry */
return (u_int16_t)sum;
}
They used to call this "wrap-around carry".
John Polstra, CDC-1604 survivor
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