Binary method for calculating hardware/Software mults/divs.

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21 comments, last by Dmytry 19 years, 2 months ago
Result =
ai*2^0 + aj*2^1 + ak*2^2 + al*2^3 + am*2^4 + an*2^5+ ao*2^5 + ap*2^7 +

bi*2^1 + bj*2^2 + bk*2^3 + bl*2^4 + bm*2^5 + bn*2^6+ bo*2^7 +

ci*2^2 + cj*2^3 + ck*2^4 + cl*2^5 + cm*2^6 + cn*2^7 +

di*2^3 + dj*2^4 + dk*2^5 + dl*2^6 + dm*2^7 +

ei*2^4 + ej*2^5 + ek*2^6 + el*2^7 +

fi*2^5 + fj*2^6 + fk*2^7 +

gi*2^6 + gj*2^7

Problem, how do i handle overflow?

For eg. ap + bo + cn + ddm + el + fk + gj> what if there all one?

From,
Nice coder
Click here to patch the mozilla IDN exploit, or click Here then type in Network.enableidn and set its value to false. Restart the browser for the patches to work.
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My problem is that this is a hardware multiplication unit, so i end up with a little bit of a problem.

I end up with a bit of a bit-counting and adding problem... [bawling]

From,
Nice coder
Click here to patch the mozilla IDN exploit, or click Here then type in Network.enableidn and set its value to false. Restart the browser for the patches to work.
i posted Lord Clickster of Clicksters into some of your threads, where you asked about adders. On bottom of that page are << and >> , and next is exactly about multiplication. Interactive diagram.

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