I am trying to print the numbers simply in the sequence i.e
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
using Loop, First i converted each number into Hexa printed it reset it to the decimal increment by 1 and then print the next until the number is equal to 9, When the number is equal to 9 i used DAA to simply the number and after rotating and shifting the number i eventually stored the result in the string.
The output is just fine till the 16, but after 16 the sequence repeats itself,
Desired output:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
Current Output 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,11,12,13,14,15
Why does it happens so ???
Here is my Code,
MOV CX,20 ;Number of Iterations
MOV DX,1
L1:
PUSH DX
ADD DX,30H
MOV AH,02H ;PRINT Content of DX
INT 21H
POP DX
ADD DX,1
CMP DX,09d ;If number is Greater than 9 jump to L2
JA L2
LOOP L1
L2:
PUSH DX
MOV AX,DX
DAA ;Convert to the Decimal
XOR AH,AH ;SET AH to 0000
ROR AX,1
ROR AX,1
ROR AX,1
ROR AX,1
SHR AH,1
SHR AH,1
SHR AH,1
SHR AH,1
ADC AX,3030h
MOV BX,OFFSET Result
MOV byte ptr[BX],5 ; Length of the String
MOV byte ptr[BX+4],'$' ;5th position of string , $=Terminator
MOV byte ptr[BX+3],AH ;2nd Number onto 4th position
MOV byte ptr[BX+2],AL ;3rd number onto 3rd Position
MOV DX,BX
ADD DX,02 ;1st 2 positions of String are type of string and
length respectively
MOV AH,09H ;to print the string
INT 21H
POP DX
ADD DX,1
LOOP L2
MOV AH,4CH ;Return control to the DOS
INT 21H
P.S: I took help from this chart in understanding the numbers.
Just giving it a try, though I'm not sure, and I can't quickly test this.
But instead of using two loops I'd recommend using one for the whole bunch of numbers.
Furthermore I have the feeling that the problem has to do with the DAA instruction, which I'm not used to, since it is not supported in 64 bit mode.
Anyway, here's what I'd do:
mov cx,20
mov al,1
mov bl,10 ; divisor
mov bp,offset Result ; no need to load this in the loop!!!
L1: mov dx,ax ; save to register, not to stack
cmp ax,09d
ja L2 ; number has two digits
add al,30h ; ASCII addend
; insert your output code here
jmp L3 ; jump over the two digit code
L2: xor ah,ah
div bl ; divides AX by ten (no rotate or shift needed)
; quotient in AL, remainder in AH (correct order for little endian)
add ax,3030h
; insert your output code here (note that the buffer/string address is loaded to BP)
L3: mov ax,dx
inc ax
loop L1
; done
If you wouldn't mind if one-digit numbers had a leading zero, it'd be even easier.
The div instruction is probably more expensive than daa plus ror plus shr, but your quad-rotate/shift will be even worse :-/
(As I said, I could not try it... leaving this open to you... if it doesn't work, just ask back.)
—
[update:
Another approach, especially to spare the div in this trivial case of digit separation, would be to add 6 to numbers greater nine (i. e. 10d = 0ah --(+6)--> 16d = 10h; this is what daa also does), then you can get along with the rotate/shift combination you used before.
Even better were to add 246, then to AX, after which you can simply use ror ax,8 (or rol — doesn't matter in this case), i. e. 10d = 0ah --(+246)--> 256d = 100h, as well 15d = 0fh --(+246)--> 261 = 105h. Rotate it to be 0001h or 0501h respectively, add 3030h, and you're done.
/update]
[update level="2"
What the fun... I actually intended to write it in the first level update, but forgot it somehow: instead of rolling by 8, or — if your TASM really doesn't support rolling by immediate — eight times rolling by one, you can of course also make use of the xchg instruction, which swaps values between registers, in this case
xchg al,ah
would do the job of swapping the contents of those two registers.
There's also a bswap instruction for reversing the byte order within a register, but it's obviously only available for registers of 32+ bits width.
/update]