Converting a code a* b * c written in C to LC3

Question

I don't know if this question is vague or lacks enough information but I was just wondering If I want to convert this line a = a * b * c written in C language to LC-3, how can I do it? Assuming that a, b and c are local variables and that the offset of a is 0, b is -1, c is -2?

I know that I can start off like this:

LDR R0, R5, #0; to load A
LDR R1, R5, #-1; to load B

Is there a limitation to the registers I can use? Can I use R2 to load C?

Edit:

         LDR R0, R5, #0; LOAD A
         LDR R1, R5, #-1; LOAD B
         LDR R2, R5, #-2; LOAD C
         AND R3, R3, #0; Sum = 0
LOOP     ADD R3, R3, R1; Sum = sum + B
         ADD R0, R0, #-1; A = A-1
         STR R0, R5, #0; SAVE TO A (a = a*b)
         BRp LOOP
         ADD R4, R4, R2; Sum1 = sum1 + C
         ADD R2, R2, #-1; C = C-1
         BRp LOOP
         STR R0, R5, #0; SAVE TO A (a = a*c = a*b*c)

You tell me. Is there a limitation to the registers you can use? — user253751, May 23 '22 at 12:45
I don't think there is, however, I'm not too sure how to work this whole 3 number multiplication out in LC3 — yowhatsup123, May 23 '22 at 13:02
Basically you have a bunch of instructions like lego bricks in front of you, and your job have to figure out how to put them together into a 3-number multiplication like a lego house. Did they give you a sheet telling you all the LC3 instructions? — user253751, May 23 '22 at 13:39
I see the LC3 has no multiply instruction. I guess they told you to use a loop (add A B times)? — user253751, May 23 '22 at 13:43

score 1 · Answer 1 · answered May 23 '22 at 14:58

1

If you're writing a whole program, which is often the case with LC-3, the only physical limit is the instruction set, so modulo that, you can use the registers as you like.

Your coursework/assignment may impose some environmental requirements, such as local variables and parameters being accessible from a frame pointer, e.g. R5, and having the stack pointer in R6. If so, then those should probably be left alone, but in a pinch you could save them, and later restore them.

If you're writing just a function that is going to be called, then you'll need to follow the calling convention. Decode the signature of the function you're implementing according to the parameter passing approach. If you want to use R7 (e.g. as a scratch register, or if you want to call another function) be aware that on entry to your function, it holds the return address, whose value is needed to return to the caller, but you can save it on the stack or in a global storage for later retrieval.

The calling convention in use should also inform which registers are call clobbered vs. call preserved. Within a function, call-clobbered registers can be used without fuss, but call-preserved registers require being saving before use, and later restored to original values before returning to the function's caller.

answered May 23 '22 at 14:58

Erik Eidt

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Oh! I understand that now! However, even after loading `c`, I am confused how the actual multiplication of the 3 values can be coded using LC3 – yowhatsup123 May 23 '22 at 15:04
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Well, that's a different question, but in short, multiply by repetitive addition is most basic approach, but if you need performance, then bit shifting, though more complex, is faster. – Erik Eidt May 23 '22 at 15:06
I have an idea of the repetitive addition, I'm thinking if I can do `a = a*b` and then `a = a*c` to make it `a = a*b*c`, is that the right logic? – yowhatsup123 May 23 '22 at 15:27
Yes that'll work. But probably don't need to actually assign back to `a` for the first one as long as you do the next multiply with that intermediate calculation (the result of a*b, then *c). – Erik Eidt May 23 '22 at 15:32
@yowhatsup123 Yes. – user253751 May 23 '22 at 15:36
So I got to `a=a*b` (attached my code as an edit in my original question), should my `a = a*c` now be after after the loop or before? – yowhatsup123 May 23 '22 at 15:42
It doesn't really matter: multiplication is commutative, so do it as you like. – Erik Eidt May 23 '22 at 15:48
There's no need to store to `a` inside the loop, that can wait until you're all done with both multiplies. – Erik Eidt May 23 '22 at 15:48
I tried writing something but I'm not sure if I've got the logic right, could you check the code out once? – yowhatsup123 May 23 '22 at 15:57
No, that's not going to work. You're trying to do two multiplies at once, when you really want to do two separate multiplies. – Erik Eidt May 23 '22 at 15:59
I have no idea how to do that, could you please provide some hint? – yowhatsup123 May 23 '22 at 16:07
Each multiplication needs to be separate, so write the code so as to complete one multiplication before starting the other. – Erik Eidt May 23 '22 at 16:11
Is my logic right now? – yowhatsup123 May 23 '22 at 16:36
I know you said we don't need to store the value in `a` before actually completing the calculation but I don't know it any other way round – yowhatsup123 May 23 '22 at 16:37
Think about what you're trying to multiply together for the 2nd multiplication. Also, clear R4 before starting the 2nd multiply. And don't try to share code between the two multiplications, they are separate, so separate them. – Erik Eidt May 23 '22 at 16:40
Also think about where the result of the first multiplication lives. – Erik Eidt May 23 '22 at 16:49
is it stored in R3? – yowhatsup123 May 23 '22 at 16:55
Yes, and that is then supposed to be one operand of the 2nd multiply. – Erik Eidt May 23 '22 at 16:56
So should be `ADD R4, R3, R2` for the second multiplication? – yowhatsup123 May 23 '22 at 16:57
No, that won't sum anything -- it will just do the same add over and over. – Erik Eidt May 23 '22 at 16:59
Else, `R3, R3, R2`? – yowhatsup123 May 23 '22 at 17:01
Ok, that will sum something, but R3 has the result of a*b; it is one of the operands to this multiplication. So, you can't really use that for the result of this next multiplication. Look at the first multiplication to see the pattern of how the two operands and the result are associated. – Erik Eidt May 23 '22 at 18:49

Converting a code a* b * c written in C to LC3

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