assembly

[xxxx123]

is this the right place to ask questions on assembly???

[whizkid2300]

Not really, but there are people here that can help you.

If you have a question ask it, and someone will get back to you.

[Negative]

Actually, this is the right place to ask

[xxxx123]

ok.my first question is regarding assembly arimethic.i understand that when u want do add a or subtract two data's that are more than a word long,u would have to process it one word at a time right?so what about multiplying and dividing data that is more than a word long???

thanks

[sec_ware]

Hi

Unfortunately, it is not such a simple task to perform what you want,
at least as much as I know.

I suggest you look up the SSE2 (or start with SSE[1])-instruction set.
I have some experience with those, but I mainly dealt with the optimisation
of single/double-precision floating point arithmetics - by parallelising rather
than having quadword operations[2]. The ops you are interested in are
paddq, psubq and pmuludq (division has to be replaced by inversion (~approximated)
and multiplication. Be careful, you might run into give carry/borrow problems.

Cheers.

[1] http://www.intel.com/software/produc...d/download.htm
[2] http://cache-www.intel.com/cd/00/00/...ng_started.pdf

[xxxx123]

thank you for the links.

what if the code has to be written on a 8088 processor which has only has 16-bit registers?i just need to know the method to accomplish this task.but a source code would be much appreciated.

i'm just asking this question for academic purposes only and i am just stumped at this point.

thanks.

[SirDice]

Make use of the carry flag.

I don't know much about i386 assembly but I used to do a lot of 6510 and 68000 assembly.
The idea should be the same...

If you add 2 numbers together the carry flag will be set when it doesn't 'fit' anymore.
Add the next 2 numbers and add the carry flag. etc... Just like you would do a 'regular' addition on paper.

If you add 9 and 2 i.e. you're basicly using a carry because 11 doesn't 'fit' anymore into 1 digit (on a 10based number system).

So you add 9 and 2... which is 1 and a carry. The next digit is 0 + 0 + carry(1). That makes 11.
Same idea works for bigger numbers..

[xxxx123]

sir dice,

what if it's a multiply or division operation?

[SirDice]

How do you multiply? Using MUL or shifting bits?

Shifting bits is probably easier and perhaps faster (MUL uses a lot of cycles and may not be available RISC i.e.).

I hope you know that shifting 1 bit to the left is the same as multiplying by 2?

I'll use a 4 bit example but the same principle works for 8, 16, 32 and 64 bits.

4 bit: 2 x 4

0100 (4) shift 1 bit to the left makes 1000 (8).

4 bit: 2 x 9

1001 (9) shift 1 bit to the left makes 0010 and the carry bit gets set. Shift that carry bit into the next 4 bits.. and you get 0001 0010 (18)

What if you need to multiply by 3?

4 bit: 3 x 4 = (2 x 4) + 4
shift 1 bit like above and add 4... 1000 + 0100 = 1100 (12)

4 bit: 3 x 9 = (2 x 9) + 9
same as above for 2x9 and add 9.. 0001 0010 + 0000 1001 = 0001 1011 (27)

Does that make sense?

Edit: Just in... You could also use a repeated addition (loop). 6x4 = 4+4+4+4+4+4

[xxxx123]

sir dice ,

thank you.that was briliant.
"3 x 4 = (2 x 4) + 4"
"3 x 9 = (2 x 9) + 9"
how did you come to that conclusion?

so i guess that for division we just use repeated subtration right or shift to the right,correct?