Intel 8085 program to multiply two integers of size 20 digits each

Also anthrax759's suggestion is correct, his method of accumulation is not the effective one. Please take a look here for multiplication algorithms:
www.cc.gatech.edu/classes/ cs3760b_99_winter/lectures/lecture5.pdf

mstrachan,

This is clearly an excercise in using the carry flag.

You can multiply two numbers together in assembler and get the carry flag set if the calculation overflowed beyond the limit of the values. All you need to do then is to use that carry while multiplying the next block up the line. You can extend multiplication as far as you like this way. This is like junior/primary school long multiplication but with much bigger numbers.


Paul

Hi,
> You can multiply two numbers together in assembler and get the carry
> flag set if the calculation overflowed beyond the limit of the values.

8085 doesn't have MUL or IMUL instructions. You have to do successive approximation only. But that method will not be suitable in any way to this large number.

Assuming 3 instructions per iteration using a 3 MHz processor, it will take about 3e13 seconds, which means about, 1000000 years

---
Harish

mstrachan, another method would be SHIFT LEFT and ADD method but it is difficult to implement for such astronomical numbers.

Many other methods are given here:
http://www.daniweb.com/tutorials/tutorial1766.html

Hi mstrachan ,
mgharish is correct,
8085 is 8-bit processor.
For storing this 19837653929987654321 (20 digits) itself will take 8 mem locations.  So it will need 1+7 inner iterations to manage count value itself. And 8 more locations for multiplier. and 8 conditions to check overflow in every addition since one register can hold only value upto 256, as per mathematical rule result may come in upto 40digits (sum of digits of both numbers) so for managing this number you need to have 16 more registers and 16 more conditions for overflow,
and nearly 16+8+8+ (8 to 16 for actual addition) addition statements you need.

So finally  as mgharish told it will take long years to complete. Using 8086 itself will take longer time. So better you can choose any 32 bit processor.

>>mgh_mgharish
>> Another method would be SHIFT LEFT ....
What I feel is, we can not use and shift mnemonics. Because we can not take this as whole value. We can take only part of (mod of 256) only. For this purpose we can not use shift mnemonics. And if you use any shift mnemonic, we will not get any carry flag to set.

This problem looks like an application oriented( or some result oriented). 8085 is designed for controlling machines (control oriented). For application oriented you need to use more than 80(1)86.

So definitely this can not be achieved using 8085. If program is developed then we will not be living to see the results.

Well you know, this is a question that is given to a whole bunch of students,
all over the world. I spoke to someone who already did the exact assignment
and got it to work correctly. It doesn't have to run on an 8085 machine, it
is going to run on an emulator that will be emulating an intel 8085 machine
not on an actual intel 8085 processor.

Maybe I forgot to mention that.

You can do it in ASCII, by following the exact method you use to do it by hand.  Exactly.

Not trivial, but not really very hard if you just stick to doing it exactly as described.

At the lowest level, to multiply two one-digit numbers, you can do it one of two ways:

(1)  Just do repetitive addition, not too slow for single digits.

(2)  Have a 10 x 10 multiplication table in memory that you just index into.

abith,

>>And if you use any shift mnemonic, we will not get any carry flag to set.
I am not familliar with 8085. Is there a ROL/ROR instruction? If so you can rotate and check the lowest bit after for carry state.

Paul

There are rotate instructions which can set the carry flag.

RAR  --> Rotate Right Through Carry
RAL  --> Rotate Left Through Carry
RRC  --> Rotate Right Directly
RLC  --> Rotate Left Directly

Through carry means (M/LSB = C; C = L/MSB)
Directly means (C = M/LSB = L/MSB)

>> mgh_mgharish
i meant for addition you can not use shift instructions, that carry flag is one of bits of acc. this will not help for any addition, since we have to do repeated addition for multiplication.

abith, I still don't understand what you mean. You can use SHIFT instructions for BOTH addition and multiplication.

mgh_mgharish , i dont know whether it is possible to add two numbers using shift instructions. if it is possible can you brief it.

You asked it for multibyte addition right ?

I was replying to this:

> For this purpose we can not use shift mnemonics.
> And if you use any shift mnemonic, we will not get any carry flag to set.

Shift instructions do set the carry flag and they can be used for higher bytes

Hello Everyone,

Thanks for your help, regarding this matter. I have read
all your messages and have finsihed the assignment thanks
to your help.

Please do not post any more replies.

Thanks,

Manny

mstrachan, aren't you forgetting to give someone a grade? Whoever pointed you in the right direction should get the points, not only because it's courteous but also to mark this question as having a solution...

anthrax759,

I apologize I am new here. I'll get on that right away.

Ah, I feel guilty now because I'm not the one who pointed you in the right direction (?) I'll try to transfer the pts to both mgharish and abith as I think they're more deserving than I...
-aNthrax

hi anthrax759 ...
we appriciate your honestness..
its ok let it be...