| --- Log opened Tue Nov 06 00:00:59 2007 |
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| 04:02 | <@ToxicFrog> | <McMartin> "Say you have a problem. Then you decide, hey -- let's use XML. Now you have two problems." |
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| 16:25 | | * AnnoDomini ponders this last assignment out of Microprocessor Technology. |
| 16:26 | <@AnnoDomini> | Given two four-digit BCD numbers A and B, we are to produce C = B + (A - B). |
| 16:27 | <@AnnoDomini> | C being a maximum of six-digits. |
| 16:27 | | * AnnoDomini ponders whether it's easier to slough through BCD arithmetic, or to translate and retranslate to and from BCD. |
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| 16:32 | <@AnnoDomini> | Hm. I might use the index register to cunningly store a four-digit number all at once, rather than having to use multiple memory cells. |
| 16:38 | <@AnnoDomini> | Though no. I can't really perform arithmetic simply enough on such numbers. Plan A it is. |
| 16:49 | <@AnnoDomini> | Gaaah. Why does this processor have logical shift right, but not logical shift LEFT?! |
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| 16:58 | | * Vornicus eyes |
| 16:58 | <@Vornicus> | I don't see how C is different from A. |
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| 17:02 | <@gnolam> | Indeed. Unless you're supposed to take some sort of overflow into account, C = A. |
| 17:03 | <@Vornicus> | And since your result is stated "maximum six digits", overflow isn't an issue either. |
| 17:05 | <@AnnoDomini> | I don't understand. Given two 8-bit accumulators, and any number of 8-bit memory cells, how would you convert a BCD number into hexadecimal? |
| 17:08 | | You're now known as TheWatcher[afk] |
| 17:08 | <@Vornicus> | Lookup tables, multiplication, and add. |
| 17:10 | <@AnnoDomini> | Multiplication? Hahahaha. No such thing. I'm lucky to have addition. |
| 17:11 | <@AnnoDomini> | (It's a Motorola 6800.) |
| 17:11 | <@Vornicus> | You need to multiply by 100. If you have shift left, it's A << 6 + A << 5 + A << 2 |
| 17:12 | <@Vornicus> | YOu should have shift left though. I can't imagine not having it, it's how you do, well, anything |
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| 17:16 | <@Vornicus> | You need to multiply by 100. If you have shift left, it's A << 6 + A << 5 + A << 2 |
| 17:16 | <@Vornicus> | YOu should have shift left though. I can't imagine not having it, it's how you do, well, anything |
| 17:18 | <@AnnoDomini> | There's arithmetic shift left. And apparently no shift operation will let me grab the C flag for the right-/left-most bit. |
| 17:19 | <@Vornicus> | blarg |
| 17:19 | <@Vornicus> | 2 registers is pain ;_; |
| 17:20 | <@AnnoDomini> | Yeah. |
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| 18:07 | | * AnnoDomini is doing this by way of C = 2*B - A. So far, so good. I think I've managed to slough through shifting the BCD number. |
| 18:11 | | * Vornicus eyes |
| 18:11 | <@Vornicus> | I don't see how that form follows the original. |
| 18:11 | <@AnnoDomini> | Er. It was supposed to be C = B + (B - A). |
| 18:12 | <@Vornicus> | aha |
| 18:12 | <@AnnoDomini> | (Though it'd be very simple to fix with S/RP.) |
| --- Log opened Tue Nov 06 19:02:39 2007 |
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| --- Log closed Wed Nov 07 00:00:37 2007 |