mulitiplication using bitwise operators

[sandy_pt_in]

Hi C guru's
can you please tell me how to do multiplication of 2 numbers using
bitwize operators?? expecting reply.

[Morris Dovey]

sa*********@yahoo.com wrote:

can you please tell me how to do multiplication of 2 numbers using
bitwize operators?? expecting reply.

To multiply:

[1] Initialize product to zero.
[2] While multiplicand is greater than zero, add (see below)
multiplier to product and decrement multiplicand.

To add:

[1] Initialize sum to augend
[2] Exclusive OR sum and addend
[3] AND addend and sum, shift this result left one position
[4] Result from [2] replaces sum
[5] Result from [3] replaces addend
[6] If addend isn't zero, continue from [2]
--
Morris Dovey
West Des Moines, Iowa USA
C links at http://www.iedu.com/c

[Keith Thompson]

sa*********@yahoo.com writes:

Hi C guru's
can you please tell me how to do multiplication of 2 numbers using
bitwize operators?? expecting reply.

Why would you want to do that?

If this is homework, we won't do it for you. If you have some other
reason to want to do this, you need to (1) explain why you can't just
use the "*" operator, and (2) show us some C code so we can help you
with it.

Off the top of my head, I'm sure it's possible, but it's probably
pretty difficult.

--
Keith Thompson (The_Other_Keith) ks*@cts.com <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://www.sdsc.edu/~kst>
Schroedinger does Shakespeare: "To be *and* not to be"

[Dave Thompson]

On Fri, 19 Sep 2003 01:32:06 -0500, Morris Dovey <mr*****@iedu.com>
wrote:

sa*********@yahoo.com wrote:

can you please tell me how to do multiplication of 2 numbers using
bitwize operators?? expecting reply.
To multiply:

For nonnegative (or unsigned) integers only. Otherwise, add more
complexity. (Well, not add, that's not bitwise. Shift in?)
[1] Initialize product to zero.
[2] While multiplicand is greater than zero, add (see below)
multiplier to product and decrement multiplicand.
Decrement is not bitwise. Change that to add ~0U.
Or to the specialized algorithm: loop changing all trailing zero bits
to one and (then) the last one to zero.

Or, even better, use "Russian peasant":
[2] While multiplicand is greater than zero,
[2A] if low bit of multiplicand is set, add multiplier to product.
[2B] (always) shift multiplier left and multiplicand right.
To add:

[1] Initialize sum to augend
[2] Exclusive OR sum and addend
[3] AND addend and sum, shift this result left one position
[4] Result from [2] replaces sum
[5] Result from [3] replaces addend
[6] If addend isn't zero, continue from [2]

You could do full-adder logic with carry-save, and only complete the
ripples at the end. But I don't think it's feasible to do look-ahead.
Or at least not worth the trouble.

drHTHtOP :-)

- David.Thompson1 at worldnet.att.net