For some reason, when I switch to mouse input switch back to keyboard input for my program, increasing and decreasing the counter has no effect. It works perfectly in the first loop where we input characters.
Here is the program, any advice?
Look at whatspeed jump for reference after mousetime jump.
CX counter is not updating or something along those lines. (I do not know if the counter is not updating or weather there is a problem occuring in comparing CX after I switch from mousetime to keytime . )
CLEAR MACRO
MOV AX,0600H
MOV BH,07
MOV CX,0000
MOV DX,184FH
INT 10H
ENDM
CURSOR MACRO Col, Row
MOV AH,02
MOV BH,00
MOV DL,Col
MOV DH,Row
INT 10H
ENDM
DISP MACRO MES
MOV AH,09
MOV DX,OFFSET MES
INT 21H
ENDM
CLEARLINE MACRO ROW
CURSOR 00,ROW
DISP SPACES
ENDM
ALWAYSONSCREENINFO MACRO
CURSOR 16,00
DISP TITLE1
CURSOR 50,00
DISP NAMES1
CURSOR 54,01
DISP NAMES2
CURSOR 33,7
DISP MENU
CURSOR 24,9
DISP OPTION1
CURSOR 24,10
DISP OPTION2
CURSOR 22,11
DISP DASHES
CURSOR 18,12
DISP MOUSEMENU
CURSOR 14,13
DISP OPTION3
CURSOR 8,14
DISP OPTION4
CURSOR 13,15
DISP CHARMENU
CURSOR 21,16
DISP KILL
ENDM
.MODEL SMALL ; RUN THE PROGRAM IN DIMENSIONS 79x24
.STACK 64H
.DATA
TITLE1 DB 'MICROPROCESSOR - EENG410','$'
NAMES1 DB 'name1 & name2','$'
NAMES2 DB 'std1 std2','$'
MENU DB 'MENU','$'
OPTION1 DB '1. Press "U" to Speed up the motor','$'
OPTION2 DB '2. Press "D" to Slow Down the motor','$'
OPTION3 DB '1. Right click the mouse to turn motor direction to clockwise','$'
OPTION4 DB '2. Left click the mouse to turn the motor direction to anti-clockwise','$'
DASHES DB '----------------------------------------','$'
MOUSEMENU DB '(Press M to switch to the mouse options)','$'
CHARMENU DB '(Press the scroll button to switch back to fist 2 options)','$'
SPEEDUP DB 'DC motor is speeding up ','$'
SLOWDOWN DB 'DC motor is slowing down','$'
RIGHT DB 'DC motor will now rotate clockwise','$'
LEFT DB 'DC motor will now rotate anti-clockwise','$'
KILL DB '-----To exit the program, press "E"-----','$'
PROGRAMEND1 DB 'Thank you for using our program','$'
PROGRAMEND2 DB 'The program has been terminated','$'
SPACES DB ' ','$'
TRY DB 'Please try again','$'
SPEEDNOW DB 'Speed:','$'
DIRECTION DB 'Motor Direction:','$'
CLOCK DB 'Right','$'
COUNTER DB 'Left ','$'
N1 DB '1','$'
N2 DB '2','$'
N3 DB '3','$'
N4 DB '4','$'
N5 DB '5','$'
N6 DB '6','$'
N7 DB '7','$'
MAXSPEED DB 'Max speed is 7','$'
MINSPEED DB 'Min speed is 1','$'
.CODE
MAIN: MOV AX,#DATA
MOV DS, AX
CLEAR
ALWAYSONSCREENINFO
MOV CX,1
CURSOR 28,21
DISP SPEEDNOW
CURSOR 35,21
DISP N1
CURSOR 22,22
DISP DIRECTION
CURSOR 39,22
DISP CLOCK
MOV CX,1
L0: MOV AH,00 ;LOOP START
INT 16H
CMP AL,' '
JE SPACED
CMP AL,'U'
JE FASTER
CMP AL,'u'
JE FASTER
CMP AL,'D'
JE SLOWER
CMP AL,'d'
JE SLOWER
CMP AL,'M'
JE MOUSETIME
CMP AL,'m'
JE MOUSETIME
CMP AL,'E'
JE EXIT
CMP AL,'e'
JE EXIT
JNE TRYAGAINKEY
A1: MOV AX,03
INT 33H
CMP BX,0
JE A1
CMP BX,1 ;MOUSE LEFT
JE MRIGHT
CMP BX,2 ;MOUSE RIGHT
JE MLEFT
CMP BX,3
JE KEYTIME
SPACED: CLEARLINE 19
JMP L0
FASTER: CLEARLINE 19
CURSOR 27,19
DISP SPEEDUP
JMP INCREASE
SLOWER: CLEARLINE 19
CURSOR 27,19
DISP SLOWDOWN
JMP DECREASE
MOUSETIME: JMP A1
KEYTIME: JMP L0
MRIGHT: CLEARLINE 19
CURSOR 18,19
DISP RIGHT
JMP TRIGHT
MLEFT: CLEARLINE 19
CURSOR 18,19
DISP LEFT
JMP TLEFT
TRYAGAINKEY: CLEARLINE 19
CURSOR 28,19
DISP TRY
JMP L0
INCREASE: CMP CX,7
JE CANNOTINCREASESPEED
INC CX
JMP WHATSPEED
DECREASE: CMP CX,1
JE CANNOTDECREASESPEED
DEC CX
JMP WHATSPEED
TRIGHT: CURSOR 39,22
DISP CLOCK
JMP A1
TLEFT: CURSOR 39,22
DISP COUNTER
JMP A1
CANNOTINCREASESPEED: CLEARLINE 19
CURSOR 27,19
DISP MAXSPEED
JMP L0
CANNOTDECREASESPEED: CLEARLINE 19
CURSOR 27,19
DISP MINSPEED
JMP L0
WHATSPEED: CMP CX,1
JE N11
CMP CX,2
JE N22
CMP CX,3
JE N33
CMP CX,4
JE N44
CMP CX,5
JE N55
CMP CX,6
JE N66
CMP CX,7
JE N77
N11: CURSOR 35,21
DISP N1
JMP L0
N22: CURSOR 35,21
DISP N2
JMP L0
N33: CURSOR 35,21
DISP N3
JMP L0
N44: CURSOR 35,21
DISP N4
JMP L0
N55: CURSOR 35,21
DISP N5
JMP L0
N66: CURSOR 35,21
DISP N6
JMP L0
N77: CURSOR 35,21
DISP N7
JMP L0
EXIT: CLEAR
CURSOR 21,12
DISP PROGRAMEND1
CURSOR 21,13
DISP PROGRAMEND2
MOV AH, 4CH
INT 21H
END MAIN
The fault is caused because the mouse interrupt 33h function AX=0003h returns the mouse position in CX and DX.
This overwrites your "counter" in register CX.
It is always a dangerous game to keep values in registers throughout a program. Better to have a memory variable location.
You could also try PUSH CX before executing the mouse functions and POP CX after, but your program structure and flow might not be conducive to that (I have not followed it fully).
Also, the reason why your program does not recover from (in your context) the absurd value of CX is because of the test instruction you used. For example here
INCREASE: CMP CX,7
JE CANNOTINCREASESPEED
you must always be cautious and trap any value that is out of range (even if you think it won't be)
JGE CANNOTINCREASESPEED
Also with
CANNOTINCREASESPEED: MOV CX,7 ;clamp the value
...
Related
iam trying to write an assembly code 8086 i have two array X AND Y size 10 and i want to enter a number and check if its positive y will be 1 if negative it will be 0
but my code just displayed what i enter a number but i want it like this example .
example:
X : 2 -5 -7 8 -4 9 11 -12 90 -50
Y : 1 0 0 1 1 0 1 1 0 1 0
I wrote:
.MODEL SMALL
.DATA
x db 10 dup ("?")
y db 10 dup ("?")
.CODE
MOV AH,1
INT 21H
mov si,offset x
mov di,offset y
mov [si],al
cmp si,0
ja l1
jmp l2
l1 : mov al , 1
mov [di],al
jmp l3
l2 : mov ah , 0
mov y[di] , al
jmp l3
l3 :
MOV AH, 2
mov dl,al
INT 21H
cmp si,10
je l4
inc si
inc di
l4 :
.EXIT
END
The list of problems is long. Check out these comments and try to see why the error exists.
<<<< add label for a 10-time repeat
MOV AH,1
INT 21H
mov si,offset x <<<< move higher up for 1-time execution
mov di,offset y <<<< move higher up for 1-time execution
mov [si],al
cmp si,0 <<<< the byte to compare is in AL
ja l1 <<<< requires a SIGNED conditional, use JGE
jmp l2 <<<< reverse condition and drop this jump
l1 : mov al , 1
mov [di],al
jmp l3
l2 : mov ah , 0 <<<< should be AL
mov y[di] , al <<<< DI already refers to the y-array, use just [DI]
jmp l3 <<<< redundant
l3 :
MOV AH, 2
mov dl,al
INT 21H
cmp si,10 <<<< 10 is wrong
je l4
inc si <<<< move above checking iterator
inc di <<<< move above checking iterator
<<<< needs jumping back in order to repeat
l4 :
Next code, just like yours, inputs a single character from the keyboard and treats its ASCII code as the byte you want to store in the x array. For negative bytes you should input extended ASCII's. Use the Alt followed by a number up to 255.
.CODE
xor bx, bx
again:
mov ah, 01h ; DOS.GetCharacter
int 21h ; -> AL
mov x[bx], al
mov ah, 255
shl ax, 1
not ah
mov y[bx], ah ; 1 for positive, 0 for negative
inc bx
cmp bx, 10
jb again
.EXIT
I am trying to solve this problem about loops. I am using a push and pop method instead of using a separate register to store data.
.model small
.stack
.code
m proc
mov ax,0b800h
mov es,ax
mov di,7d0h
mov ah,7 ; normal attribute
mov al,'A'
mov cx,5
x: stosw
push ax ;mov dl,al ; dl='A'
push di
mov al,'1'
stosw
pop di
add di,158
pop ax ;mov al,dl
inc al
loop x
mov ah,4ch
int 21h
m endp
end m
I am unable to loop the mov al, '1'.
The output should be like this:
A1
B2
C3
D4
E5
Can anyone show the correct code? Thank you.
Consider the ASCII codes involved:
Letter Digit Difference
A1 65 49 16
B2 66 50 16
C3 67 51 16
D4 68 52 16
E5 69 53 16
See how the difference is always 16 ? That's what next solution exploits:
...
mov ax, 0700h + 'A' ; WhiteOnBlack 'A'
x: stosw ; Stores one letter from {A, B, C, D, E}
sub al, 'A' - '1' ; Convert from letter to digit
stosw ; Stores one digit from {1, 2, 3, 4, 5}
add al, 'A' - '1' + 1 ; Restore AL and at the same time increment
add di, 160 - 4 ; Move down on the screen
cmp al, 'E'
jbe x
...
You don't always need to use CX and the LOOP instruction to work with a loop. Anyway the LOOP instruction is to be avoided for speed reasons!
Here is my code:
data segment
gio db 1,2,3,4,5,6
ricxvi db 1
jami db 0
x db ?
ends
stack segment
db 128 dup(0)
ends
code segment
start:
MOV AX,DATA
MOV DS,AX
mov cx, 6
lea si, gio
mov ah, 0
n1:
mov al, [si]
cmp ricxvi, 3
je n2
jmp n3
n2:
add jami, al
mov ricxvi, 1
jmp n4
n3:
add ricxvi, 1
push ax
n4:
add si, 1
add di, 1
loop n1
mov ricxvi, 1
mov ax, 0
mov cx, 6
lea si, gio
n5:
cmp ricxvi, 3
je n6
jmp n7
n6:
mov ricxvi, 1
add si, 1
loop n5
n7:
pop [si]
add si, 1
loop n5
mov ax, 4c00h
int 21h
ends
end start
I have a array named gio And I'm trying to reverse this array but leave every 3th element on its position. Meaning I want to get the output like this 5,4,3,2,1,6 but As I inspect variables, in array I have 5,4,2,1,B8. I have noticed that when program first hits pop [si] whole array changes, exploring variables shows me that its 5, NULL, 3, 4, 5, 6 should it not be 5,2,3,4,5,6? I'm using emu8086. Question may sound silly as I'm new with assembly. Thanks.
There are three errors :
You are poping two bytes into [si] but you only need one byte. The solution is to pop two bytes into a register and move one byte into [si].
Under label n6 you got loop n5, but when cx becomes zero the loop doesn't jump and the block n7 is executed when it shouldn't.
You forgot to increase the "3" counter under label n7.
Here are the fixes :
n6:
mov ricxvi, 1
add si, 1
loop n5
jmp finale ;◄■■ SKIP NEXT BLOCK WHEN CX=0.
n7:
add ricxvi, 1 ;◄■■ INCREASE THE "3" COUNTER.
pop ax ;◄■■ POP TWO BYTES.
mov [si], al ;◄■■ USE ONE BYTE ONLY.
add si, 1
loop n5
finale: ;◄■■ END LABEL.
mov ax, 4c00h
int 21h
I'm trying hard to study Assembly language.
I really need help to store different 10 items in an array
I want to build a program that will accept items from 10 choices and it will store it.
Those 10 items have different value.
Example
Item 1 = 10$
Item 2 = 4$
So that when the User tries to choose Item1 and Item2 it will show the
sum of both items.
I'll be gladly be happy if someone can share his/her own code that can store 10 items with the sum of all items. Thanks
Here's my code:
_start:
mov eax,3 ;number bytes to be summed
mov ebx,0 ;EBX will store the sum
mov ecx, x ;ECX will point to the current element to be summed
top: add ebx, [ecx]
add ecx,1 ;move pointer to next element
dec eax ;decrement counter
jnz top ;if counter not 0, then loop again
done:
add ebx, '0'
mov [sum], ebx ;done, store result in "sum"
display:
mov edx,1 ;message length
mov ecx, sum ;message to write
mov ebx, 1 ;file descriptor (stdout)
mov eax, 4 ;system call number (sys_write)
int 0x80 ;call kernel
mov eax, 1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
global x
x:
db 2
db 4
db 3
sum:
db 0
My code here does have problem.
x:
db 2
db 4
db 3
sum:
db 0
Since you've defined all of your variables to be of byte type, you must process them as bytes! You're reading and writing dwords in your program.
This could have been OK:
top: add bl, [ecx]
add ecx, 1 ;move pointer to next element
dec eax ;decrement counter
jnz top ;if counter not 0, then loop again
done:
add bl, '0'
mov [sum], bl ;done, store result in "sum"
I am having problem on how exactly should i use loop to get the desire output in this program,
What i want to do is to take input any number from the user and then sort that number in descending order,
I tried my best here to explain every step of the code in the comment.
here is my code,
STSEG SEGMENT
DB 64 DUP(?)
STSEG ENDS
DTSEG SEGMENT
SNAME DB 24 DUP("$")
DTSEG ENDS
CDSEG SEGMENT
MAIN PROC
ASSUME CS:CDSEG, DS:DTSEG, SS:STSEG
MOV AX,DTSEG
MOV DS,AX
MOV ES, AX ;ES:DI
MOV DX, OFFSET STRNG1
MOV AH,09
INT 21H
XOR DX,DX
MOV BYTE PTR SNAME, 40
MOV DX, OFFSET SNAME
MOV AH, 0AH
INT 21H
PUSH DX ;Hold the input number in a stack until we clear the screen and set the cursor
; The clear screen and cursor position code is here which i didn't really mention.
;What we need to do now is to compare first number to each other number and store the greatest
of two on first position.
MOV BX,DX ;Copy un-sorted number to BX,
MOV AX,BX[1] ;the length of the number which is stored on the first position of the string
XOR AH,AH ;Empty AH
MOV CL,AL ;MOVE AL into CL for looping 6 times
SUB CL,1
MOV SI,02H ;the number is stored in string array from the 2nd position
;Suppose the input number is the following,
;[6][3][9][1][8][2][6]
;this is how it should work,
; Loop 6 times , CX = 6
[7][6][3][9][1][8][2][6] ; 7 is length of the number which is already copied in CX above.
; but we need 6 iterations this is why we subtract 1 from CL above.
; 6 > 3 ?
; Yes, then BX[SI] = 6 and BX[SI+1] = 3
; 6 > 9 ?
; NO, then BX[SI] = 9 and BX[SI+2] = 6
; 9 > 1
; Yes, then BX[SI] = 9 and BX[SI+3] = 1
; 9 > 8
; Yes, then BX[SI] = 9 and BX[SI+4] = 8
; 9 > 2
; Yes, then BX[SI] = 9 and BX[SI+5] = 2
; 9 > 6
; Yes, then BX[SI] = 9 and BX[SI+6] = 6
; After first iteration the incomplete sorted number is,
;[9][3][6][1][8][2][6]
;Similarly here i need to loop 5 times now by comparing the 2nd number which is a 3 with all ;the number after it and then loop 4 times then 3 times then two times and then 1 time.
L1:
;Loop 1 must iterate 6 time for the supposed input number,
;but i couldn't be able to write the proper code as i always get out of registers. kindly help me out
L2:
LOOP L2
Loop L1
Kindly help me with the nested loop where i've stucked.
The loop instruction uses the cx register.
So you must either preserve cx for the outer loop with e.g. push cx (before L2:) and pop cx (after loop L2):
mov cx,5
L1:
push cx
mov cx,6
L2:
. . . do stuff inner
loop L2
pop cx
. . . do stuff outer
loop L1
or remember that loop is roughly equal to dec cx jnz, so do e.g.
mov dx,5
L1:
mov cx,6
L2:
... do stuf inner
loop L2
.. do stuff outer
dec dx
jnz L1
Possible off-by-one errors are intended and meant as an exercise for the reader :-)