For example, a user inputs 01010101, and the program splits it into 01 01 01 01 to read as separate values assigned to a specific constant.
I'm aware that shifts/pulling values out of a 32-bit register and masking needs to occur, though I don't know how to do so.
Use AND to mask out bit and SHR to shift bit to right
MOV dl, 0b01010101
AND dl, 0b00000001 ;dl now equals 1
MOV al, 0b01010101
AND al, 0b00000100 ;al now equals 0b00000100 = 4
SHR al, 2 ;al now equals 1
You can use bit operator "AND" to do this.
Assume now you store the original "01011100" in register AH and you want to split it into "01", "01", "11", "00" and store each of them in BH, BL, CH, CL.
You can do it like this:
MOV BH AH ;now the BH is 01011100 ,just like AH
AND BH 0B11000000 ;since "a AND b" equals 1 only when both a and b equal 1 ,so the first two bit of AH will be the same as that of BH, and the other six bit will be 0 because one of the operand is 0 . In this way you can split it.
And you can do the rest in the same way :
MOV BL AH ;now the BL is 01011100 ,just like AH
AND BL 0B00110000 ;now BL is 00010000
MOV CH AH ;now the CH is 01011100 ,just like AH
AND CH 0B00001100 ;now the CH is 00001100
MOV CL AH ;now the CL is 01011100 ,just like AH
AND CL 0B00000011 ;now the DH is 00000000
Related
So I'm creating a simple 8086 program that calculates the sum of two two digit numbers:
.model small
.data
first db 0
second db 0
sum db 0
.stack
.code
mov ax, data
mov ds, ax
mov es, ax
mov ah, 1
int 21h
sub al, 48d
mov bl, 10
mul bl
mov first, al
mov ah, 1
int 21h
sub al, 48d
add first, al
mov ah, 1
int 21h
sub al, 48d
mov bl, 10
mul bl
mov second, al
mov ah, 1
int 21h
sub al, 48d
add second, al
mov al, first
add al, second
mov sum, al
However when I check the value of the numbers specifically the sum it gives me a negative number as it's signed value. For example if the first number is 99 and the second one is 98 the sum is -59.
Does anyone have an answers that might fix this problem?
Since you used a byte for each number, every number can have 2^8 = 256 different values. So signed numbers must be between -128 and 127. When addition result get larger than 127, it represent it as corresponding negative number. You should use word(16-bit) or more to solve this problem.
I have to prepare program for 8086 processor which converts quaternary to octal number.
My idea:
Multiplying every digit by exponents of 4 and add to register. Later check the highest exponent of 8 not higher than sum from first step.
Divide by exponents of 8 until remainder equals 0. Every result of dividing is one digit in octal.
But for 16-digits number last exponent of 4 is 4^15. I suppose It isn't optimal algorithm.
Is there any other way? Maybe to binary and group by 3 digits.
Turns out you can indeed process values 3 digits at a time. Done this way, you can process strings of arbitrary length, without being limited by the size of a register. Not sure why you might need to, unless aliens try to communicate with us using ascii strings of quaternary digits with huge lengths. Might happen.
It's possible to do the translation either way (Right to Left or Left to Right). However, there's a bit of a challenge to either:
If you are processing RtL, you need to know the length of the output string before you start (so that you know where to write the digits as you compute them). This is do-able, but a bit tricky. In simplest terms, the length is ((strlen(Q) + 2) / 3) * 2. That almost gets it. However, you can end up with a blank space at the beginning for a number of cases. "1" as well as "10" will give the blank space. "20" won't. The correct value can be computed, but it's annoying.
Likewise, processing LtR has a similar problem. You don't have the problem of figuring out where to write digits, but consider: If the string to convert it "123", then the conversion is simple (33 octal). But what if you start processing, and the complete string is "1231" (155 octal)? In that case what you need to process it like "001231" (01 55). IOW, digits can be processed in groups of 3, but you need to handle the initial case where the number of digits doesn't evenly divide by 3.
Posting solutions to homework is usually something I avoid. However I doubt you are going to turn this in as your 'solution,' and it's (barely) possible that google might send someone here who needs something similar.
A few things to note:
This code is intended to be called from C using Microsoft's fastcall (it made testing easier) and compiled with masm.
While it is written in 32bit (my environment), there's nothing that particularly requires 32bit in it. Since you said you were targeting 8086, I've tried to avoid any 'advanced' instructions. Converting to 16bit or even 64bit should not present much of a challenge.
It processes from left to right.
As with any well-written routine, it validates its parameters. It outputs a zero length string on error, such as invalid digits in the input string.
It will crash if the output buffer is NULL. I suppose I could return a bool on error (currently returns void), but, well, I didn't.
I'm sure the code could be tighter (couldn't it always?), but for "homework project quality," it seems reasonable.
Other than that, that comments should explain the code.
.386
.model flat
.code
; Call from C via:
; extern "C" void __fastcall PrintOct(const char *pQuat, char *pOct);
; On Entry:
; ecx: pQuat
; edx: pOct
; On Exit:
; eax, ecx, edx clobbered
; all others preserved
; If pOct is zero bytes long, an error occurred (probably invalid digits)
#PrintOct#8 PROC
; -----------------------
; If pOct is NULL, there's nothing we can do
test edx, edx
jz Failed
; -----------------------
; Save the registers we modify (except for
; eax, edx and ecx which we treat as scratch).
push esi
push ebx
push edi
mov esi, ecx
mov edi, edx
xor ebx, ebx
; -----------------------
; esi: pQuat
; edi: pOct
; ebx: zero (because we use lea)
; ecx: temp pointer to pQuat
; Reject NULL pQuat
test esi, esi
jz WriteNull
; -----------------------
; Reject 0 length pQuat
mov bl, BYTE PTR [esi]
test bl, bl
jz WriteNull
; -----------------------
; How many chars in pQuat?
mov dl, bl ; bl is first digit as ascii. Preserve it.
CountLoop:
inc ecx ; One more valid char
; While we're counting, check for invalid digits
cmp dl, '0'
jl WriteNull
cmp dl, '3'
jg WriteNull
mov dl, BYTE PTR [ecx] ; Read the next char
test dl, dl ; End of string?
jnz CountLoop
sub ecx, esi
; -----------------------
; At this point, there is at least 1 valid digit, and
; ecx contains # digits
; bl still contains first digit as ascii
; Normally we process 3 digits at a time. But the number of
; digits to process might not be an even multiple of 3.
; This code finds the 'remainder' when dividing ecx by 3.
; It might seem like you could just use 'div' (and you can),
; but 'div' is so insanely expensive, that doing all these
; lines is *still* cheaper than a single div.
mov eax, ecx
mov edx, 0AAAAAAABh
mul edx
shr edx, 1
lea edx, [edx+edx*2]
sub ecx, edx ; This gives us the remainder (0-2).
; If the remainder is zero, use the normal 3 digit load
jz LoadTriplet
; -----------------------
; Build a triplet from however many leading 'odd' digits
; there are (1 or 2). Result is in al.
lea eax, DWORD PTR [ebx-48] ; This get us the first digit
; If there was only 1 digit, don't try to load 2
cmp cl, 1
je OneDigit
; Load the other digit
shl al, 2
mov bl, BYTE PTR [esi+1]
sub bl, 48
or al, bl
OneDigit:
add esi, ecx ; Update our pQuat pointer
jmp ProcessDigits
; -----------------------
; Build a triplet from the next 3 digits.
; Result is in al.
; bl contains the first digit as ascii
LoadTriplet:
lea eax, DWORD PTR [ebx-48]
shl al, 4 ; Make room for the other 2 digits.
; Second digit
mov cl, BYTE PTR [esi+1]
sub cl, '0'
shl cl, 2
or al, cl
; Third digit
mov bl, BYTE PTR [esi+2]
sub bl, '0'
or al, bl
add esi, 3 ; Update our pQuat pointer
; -----------------------
; At this point
; al: Triplet
; ch: DigitWritten (initially zeroed when computing remainder)
ProcessDigits:
mov dl, al
shr al, 3 ; left digit
and dl, 7 ; right digit
; If we haven't written any digits, and we are
; about to write a zero, skip it. This deals
; with both "000123" and "2" (due to OneDigit,
; the 'left digit' might be zero).
; If we haven't written any digits yet (ch == 0), and the
; value we are are about to write is zero (al == 0), skip
; the write.
or ch, al
jz Skip1
add al, '0' ; Convert to ascii
mov BYTE PTR [edi], al ; Write a digit
inc edi ; Update pointer to output buffer
jmp Skip1a ; No need to check again
Skip1:
or ch, dl ; Both check and update DigitWritten
jz Skip2
Skip1a:
add dl, '0' ; Convert to ascii
mov BYTE PTR [edi], dl ; Write a digit
inc edi ; Update pointer to output buffer
Skip2:
; Load the next digit.
mov bl, BYTE PTR [esi]
test bl, bl
jnz LoadTriplet
; -----------------------
; All digits processed. We know there is at least 1 valid digit
; (checked on entry), so if we never wrote anything, the value
; must have been zero. Since we skipped it to avoid
; unnecessary preceding zeros, deal with it now.
test ch, ch
jne WriteNull
mov BYTE PTR [edi], '0'
inc edi
; -----------------------
; Write the trailing NULL. Note that if the returned string is
; 0 bytes long, an error occurred (probably invalid digits).
WriteNull:
mov BYTE PTR [edi], 0
; -----------------------
; Cleanup
pop edi
pop ebx
pop esi
Failed:
ret
#PrintOct#8 ENDP
end
I've run a string with 1,000,000,000 quaternary digit thru it as well as all the values from 0-4,294,967,295. Seems to work.
I for one welcome our new 4-digited alien overlords.
How can I add 2 numbers that their value is on base 16 and make the result on "base 10" on assembler. For example:
"5h+5h=10h" - I know it's wrong, I just want it to be visually 10h
And not:
5h+5h=Ah
CODE:
MOV AX,5h
MOV BX,5h
ADD AX,BX
result: ax=Ah - Not the result that i want...
result: ax=10h - The result that i want.
I tried to figure it out with google but didn't find anything that can help me...
Here is the code you are looking for
MOV AX,5h
MOV BX,5h
ADD AX,BX
DAA
Now AX contains 10h
Ok so i figure it out with #Fifoernik 's help
so the problem is that if i want to do it with 16bit (for example 99h+1h) values i need to do it like this using DAA operan and CF flag
pop ax
pop bx
add al,bl
daa ; dec id values
mov cl,al
mov al,ah
jc Carry; do i have carry
add al,bh
daa ; do the magic thing
JMP finito
Carry:
add al,1 ; add the carring...
add al,bh
daa ;can some one tell me what exactly daa does?
finito:
mov ch,al
push cx
ret
daa working only on AL so you'ill need to use the carry flag to add the carry like:
AH AL
1 <----- carry
00 99 <-- DAA take care of the 99 and make it 0 when its A0h
00 01+
-- --
01 00 ---> result 100h
I've tried reading about this all over the internet but here it is my problem. I am given a string of doublewords.
I have to order in decreasing order the string of the low words (least significant) from these doublewords. The high words remain unchanged.
For ex: strin DD 12345678h 1256ABCDh, 12AB4344h
the result would be 1234ABCDh, 12565678h, 12AB4344h.
Now I tried my best writting some code but it's not working properly, my insertion procedure. If you could take a look and tell me what I'm doing wrong, I'd be greatful.
I tried running it in td mode but I just can't figure out.
assume cs:code, ds:data
data segment
s dd 12345678h, 1256ABCDh, 12AB4344h
ls equ ($-s)/4 ;this is supposed to be the length of my source string
d dd ls dup (?) ;this is my destination string
aux dw ?
aux2 dw ?
data ends
code segment
insert proc
push di ;here I use the stack to get more free registers
push cx
cmp di, offset d ;if di=offset d it means that I didn't store any number yet
je addPrim
std ;we plan on working form right to left on the string for the next part
mov cx, di
sub cx, offset d ;here I find out with how many words I have to compare the word from AX
dec di
dec di ;since I work with doublewords, for some reason I thought I should decrease di
dec di ;3 times but here my procedure gets fuzzy and doesn't work properly anymore
repeta1: ;this repeat is supposed to compare the word from AX with the rest of the least
scasw ;significant words from es:di
jge DIplus2 ;if my number from AX is bigger or equal than what's in es:di, I increment
;di twice and store it
mov bx, word ptr es:[di+1] ;this part is supposed to interchange words but it's not
;working how I planned so I don't know how to change it
mov word ptr es:[di+2], bx
loop repeta1
jmp DIplus1
DIplus2:
inc di
DIplus1:
inc di
addPrim: ;this label just adds the first word in the destination string
stosw
pop cx
pop di
inc di
inc di
cld
ret
insert endp
start:
mov ax, data
mov ds, ax
mov es, ax
mov si, offset s
mov di, offset d
mov cx, ls ; store in cx the length of the strings
jcxz exit
repeta:
lodsw ;because of little endian, my first word will be my least significant word in the
;in the doubleword so right after it is moved in ax, i apply the procedure insert
call insert
lodsw ;here it moves in ax my most significan word in the dd, so i auto store it
stosw ;in my destination string
loop repeta
exit:
mov ax, 4c00h
int 21h
code ends
end start
ls equ ($-s)/4 ;this is supposed to be the length of my source string
This actually calculates the number of elements.
mov cx, di
sub cx, offset d ;here I find out with how many words ...
At the second invocation of your insert proc this will set CX=4 which is too big given a list of only 3 values. I suggest you divide CX by 4.
dec di
dec di ;since I work with doublewords...
dec di ;3 times but here my procedure gets fuzzy
This is certainly wrong. SCASW indicates you either decrement by 4 or not decrement at all!
mov bx, word ptr es:[di+1] ;this part is supposed to interchange words...
mov word ptr es:[di+2], bx
This cannot work since the offsets are only 1 byte apart!
jmp DIplus1
This yields an single increment of DI and thus an error because you want to store a word at that spot.
I have an array of nine names:
.model tiny
.data
vardas1 db "Rokas",0ah,'$'
vardas2 db "Tomas",0ah,'$'
vardas3 db "Matas",0ah,'$'
vardas4 db "Domas",0ah,'$'
vardas5 db "Augis",0ah,'$'
vardas6 db "Vofka",0ah,'$'
vardas7 db "Marka",0ah,'$'
vardas8 db "Auris",0ah,'$'
vardas9 db "Edvis",0ah,'$'
vardai dw offset vardas1, offset vardas2, offset vardas3, offset vardas4, offset vardas5, offset vardas6, offset vardas7, offset vardas8, offset vardas9
.code
org 100h
I need to read a digit from keyboard, and then I need to print that name. For example I will push 5, and console should write "Augis". BTW, second code block aren't all code, just loop that doesn't work
paieska:
mov dx, offset _comment1 ; Just string name asking user to input digit
mov ah, 9
int 21h
mov j, 00h ; Trying to input the digit from keyboard
mov ah, 01h
mov dl, 0ah
int 21h
mov bx, offset vardai ; Add array "names" to bx register
add bx, cx ; Add cx for indexing
mov dx, [bx] ; Add first array element to dx register
add cx, 2 ; Increasing cx by 2, because I'm using data word not data byte
mov ah, 9 ; Try to print it
int 21h
cmp cx, j ; Try to compare cx (index of array) to mine inputed digit "j"
jne paieska
je end
mov ah, 01h
mov dl, 0ah ;NO NEED FOR THIS - INT21/01 DOES NOT USE DL
int 21h
MOV AH, '1' ; MIN INPUT CHAR
mov bx, offset vardai ; Add array "names" to bx register WELL, ASSIGN ACTUALLY
MOV CX,2 ;NUMBER OF BYTES TO ADD (WORDS, NOT BYTES)
LOOPN:
mov dx, [bx] ; name-pointer array element to dx register
CMP AH,AL ; MATCHING char?
JE PNAME ; YES, PRINT NAME
add bx, cx ; Add cx=2 for next name
inc AH ; next possible character input
CMP AH,'9'+1 ; allowed is '1'..'9'
jne loopn ; in allowed range
; input not 1..9
mov dx, offset errormessage
PNAME:
mov ah, 9 ; Try to print it
int 21h
jmp end
Well, I tried to edit your approach with CAPS, but it became too complicated.
Essentially, you are reading a character from the keyboard using function 01. This character arrives in AL. If all goes well, it should be '1'..'9'. Notice these are the ASCII characters '1'..'9', that is hex 31..39
Next step is to set BX to the start of the table, AH to the minimum character you anticipate and CX to 2 because the table contains words, not bytes.
Now we have a loop. Load X from the table, and check whether AL is equal to AH. If the user input 1, these will be equal, so go print the string.
Otherwise, add 2 to BX to point to the next entry in the table (this could have been done by ADD BX,2 or INC BX INC BX which would mean the MOV CX,2 would be unnecessary - just the way I wrote it...) and increment the '1' in AH to '2'.
The end-condition for the loop is when AH gets incremented from '9' to - well, ':' or '9'+1. If it hasn't reached that end-condition, then run around the loop until all of the values '1'..'9' have been tested. If you haven't got to PNAME yet, then there's an error because the character input wasn't allowed, so point to an error message and then print it.
Now jumping to the end - probably you'd want to terminate the program, so you'd execute
MOV AH,4CH
INT 21H