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SDK6800 Assembly Program Help

Hi,

I've got a question about using the 6800 emulator to create a program that multiplies 2 arbitary 8 bit numbers to generate an 8 bit number. now, i've attempted it, this is what i've done in the quote below.

Quote:

eleven .equ 11
fourteen .equ 14
P .equ 0

ldab #4

loop cba
ldaa #eleven
tst 0,x
jmp PnB
PnB ldaa #fourteen
adaa #P
jmp plus
plus asla
staa P
jmp shiftP
shiftP ldaa #P
asla
decb
tstb
bne loop
.org $0120

Okay, the teacher went through it in class, and he showed how it was shifting each step, eg:

C P A
0 0000 1011
N=1 0 1110 1011 (LSB = 1, so P+B=P)
0111 0101 (Shift Bits Right)
N=2 1 0101 0101 (LSB = 1, so P+B=P)
0 1010 1010 (Shift Bits Right)

So at first i couldn't do it, but i attempted it and i did that in the quote up the top. I don't think it's correct though cause when i look at it, it just doesn't look right. I was wondering if anyone could give me some pointers on what im doing wrong so I at least know how to fix it.

Thanks,

Ashley

Re: SDK6800 Assembly Program Help

right it didn't list the codes very well, i'll list them again:

Code:

eleven         .equ         11

fourteen         .equ         14

P         .equ         0



        ldab         #4



loop         cba

        ldaa         #eleven

        tst         0,x

        jmp         PnB

PnB         ldaa         #fourteen

        adda         #P

        jmp         plus

plus         asla

        staa         P

        jmp         shiftP

shiftP         ldaa         #P

        asla

        decb

        tstb

        bne loop

        .org $0120 



        .end

and the other code

Code:

        C         P         A

        0         0000         1011

N=1        0         1110         1011        (LSB = 1, so P+B=P)

                0111         0101         (Shift Bits Right)

N=2         1         0101         0101         (LSB = 1, so P+B=P)

        0         1010         1010         (Shift Bits Right)

Re: SDK6800 Assembly Program Help

Hi,

I've got a question about using the 6800 emulator to create a program which counts the number of factors of an input variable. The input is stored at the end of the program in the variable n. The output should be stored in a variable called numFactors. Now, i've attempted it, this is what i've done in the quote below.

ldaa n
ldab x
startloop suba numFactors
tsta
dec x
bmi stop
bra startloop
n .byte 15
x .byte 1
numFactors .byte 1
stop .end