I want to put numbers from 0 - 9 to memory cells 400h to 409h.
So for example at 400h -> 0 (put 0) and at 401h -> 1 (put 1) ..... 409h (put 9).
This is my code so far: (I dont know if it works)
IDEAL
MODEL small
STACK 100h
DATASEG
;----------
;----------
CODESEG
start:
mov ax , #data
mov ds , ax
mov es, ax
;----------
mov si , 400h
mov cx , 10
mov al , 0
agian:
mov [si],al
inc si
inc al
loop agian
;--------
exit:
mov ax,4c00h
int 21h
END start
There's a very simple way to see if your program works. Just write the values in the video memory. That way you'll know if it works.
start:
mov ax, 0B800h ;NEW
mov ds, ax
mov es, ax
;----------
mov si, 400h
mov cx, 10
mov al, 48 ;NEW value 0 -> character 0
agian:
mov [si], al
add si, 2 ;NEW 1 character occupies 2 bytes in video memory
inc al
loop agian
mov ah,00h ;NEW wait for a keystroke so you can actually see
int 16h ;NEW ... the output
If you can invest the time you could learn to use the DOS utility DEBUG.EXE. Amongst other things it allows you to single step your program and view memory .
The easiest way to check if your ASM code is working the way you expect is to run it in a debugger. If you're running on Windows, OllyDbg 2 would be a good candidate — it will show you the current values of the registers, state of the stack, etc., so you can see how they change as you step through your code. You can modify the code from inside OllyDbg too.
You can write breakpoints in your code with the int 3 instruction, or use the debugger to place breakpoints at runtime.
Related
I'm trying to create a line-drawing algorithm in assembly (more specifically Bresenham's line algorithm). After trying an implementation of this algorithm, the algorithm fails to work properly even though I almost exactly replicated the plotLineLow() function from this Wikipedia page.
It should be drawing a line between 2 points, but when I test it, it draws points in random places in the window. I really don't know what could be going wrong because debugging in assembly is difficult.
I'm using NASM to convert the program to binary, and I run the binary in QEMU.
[bits 16] ; 16-bit mode
[org 0x7c00] ; memory origin
section .text ; code segmant
global _start ; tells the kernal where to begin the program
_start: ; where to start the program
; main
call cls ; clears the screen
update: ; main loop
mov cx, 0x0101 ; line pos 1
mov bx, 0x0115 ; line pos 2
call line ; draws the line
jmp update ; jumps to the start of the loop
; functions
cls: ; function to clear the screen
mov ah, 0x00 ; set video mode
mov al, 0x03 ; text mode (80x25 16 colours)
int 0x10 ; BIOS interrupt
ret ; returns to where it was called
point: ; function to draw a dot at a certain point (dx)
mov bx, 0x00ff ; clears the bx register and sets color
mov cx, 0x0001 ; clears the cx register and sets print times
mov ah, 0x02 ; set cursor position
int 0x10 ; BIOS interrupt
mov ah, 0x09 ; write character
mov al, ' ' ; character to write
int 0x10 ; BIOS interrupt
ret ; returns to where it was called
line: ; function to draw a line at two points (bx, cx)
push cx ; saves cx for later
push bx ; saves bx for later
sub bh, ch ; gets the value of dx
mov [dx_L], bh ; puts it into dx
sub bl, cl ; gets the value of dy
mov [dy_L], bl ; puts it into dy
mov byte [yi_L], 1 ; puts 1 into yi (positive slope)
cmp byte [dy_L], 0 ; checks if the slope is negative
jl .negative_y ; jumps to the corresponding sub-label
jmp .after_negative_y ; if not, jump to after the if
.negative_y: ; if statement destination
mov byte [yi_L], -1 ; sets yi to -1 (negative slope)
neg byte [dy_L] ; makes dy negative as well
.after_negative_y: ; else statement destination
mov ah, [dy_L] ; moves dy_L into a temporary register
add ah, ah ; multiplies it by 2
sub ah, [dx_L] ; subtracts dx from that
mov [D_L], ah ; moves the value into D
pop bx ; pops bx to take a value off
mov [y_L], bh ; moves the variable into the output
pop cx ; pops the stack back into cx
mov ah, bh ; moves x0 into ah
mov al, ch ; moves x1 into al
.loop_x: ; loop to go through every x iteration
mov dh, ah ; moves the iteration count into dh
mov dl, [y_L] ; moves the y value into dl to be plotted
call point ; calls the point function
cmp byte [D_L], 0 ; compares d to 0
jg .greater_y ; if greater, jumps to the if statement
jmp .else_greater_y ; if less, jumps to the else statement
mov bh, [dy_L] ; moves dy into a temporary register
.greater_y: ; if label
mov bl, [yi_L] ; moves yi into a temporary register
add [y_L], bl ; increments y by the slope
sub bh, [dx_L] ; dy and dx
add bh, bh ; multiplies bh by 2
add [D_L], bh ; adds bh to D
jmp .after_greater_y ; jumps to after the if statement
.else_greater_y: ; else label
add bh, bh ; multiplies bh by 2
add [D_L], bh ; adds bh to D
.after_greater_y: ; after teh if statement
inc ah ; increments the loop variable
cmp ah, al ; checks to see if the loop should end
je .end_loop_x ; if it ended jump to the end of teh loop
jmp .loop_x ; if not, jump back to the start of the loop
.end_loop_x: ; place to send the program when the loop ends
ret ; returns to where it was called
section .data ; data segmant
dx_L: db 0 ; used for drawing lines
dy_L: db 0 ; ^
yi_L: db 0 ; ^
xi_L: db 0 ; ^
D_L: db 0 ; ^
y_L: db 0 ; ^
x_L: db 0 ; ^
section .text ; code segmant
; boot the OS
times 510-($-$$) db 0 ; fills up bootloader space with empty bytess
db 0x55, 0xaa ; defines the bootloader bytes
I see no video mode
just 80x25 text VGA mode (mode = 3) you set at start with cls so how can you render points? You should set the video mode you want assuming VGA or VESA/VBE see the link above.
why to heck use VGA BIOS for point rendering?
that will be slooooooooow and I have no idea what it does when no gfx mode is present. You can render points by direct access to VRAM (at segment A000h) Ideal use 8/16/24/32bit video modes as they have pixels aligned to BYTEs ... my favorite is 320x200x256c (mode = 19) as it fits into 64K segment so no paging is needed and pixels are Bytes.
In case you are using characters instead of pixels then you still can use access to VRAM in the same way just use segment B800h and chars are 16 bit (color and ASCII).
integer DDA is faster then Bresenham on x86 CPUs since 80386
I do not code in NASM for around 2 decades and closest thing to line I found in my archive is this:
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
line: pusha ;ax=x0,bx=x1,dl=y0,dh=y1,cl=col
push ax ;expecting ES = A000h
mov si,bx
sub si,ax
sub ah,ah
mov al,dl
mov bx,ax
mov al,dh
sub ax,bx
mov di,ax
mov ax,320
sub dh,dh
mul dx
pop bx
add ax,bx
mov bp,ax
mov ax,1
mov bx,320
cmp si,32768
jb .r0
neg si
neg ax
.r0: cmp di,32768
jb .r1
neg di
neg bx
.r1: cmp si,di
ja .r2
xchg ax,bx
xchg si,di
.r2: mov [.ct],si
.l0: mov [es:bp],cl
add bp,ax
sub dx,di
jnc .r3
add dx,si
add bp,bx
.r3: dec word [.ct]
jnz .l0
popa
ret
.ct: dw 0
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
So you have something to cross check (took me a while to find it in my archives as I coded whole 3D polygonal engines with textures at that time so I do not have much 2D code in NASM...)
The example expects 320x200x256c VGA video mode
If you were writing a DOS .com file, things would be a bit simpler: segment registers would all be set the same, with your code/data at offset 100 from them. And you could end with ret.
[BITS 16]
[ORG 100h]
[SEGMENT .text]
ret
As #bitRAKE and #PeterCoders pointed out in case You run this in BOOT SECTOR the org is ok. However in such case there is no OS present so if you were going to do more with the stack or any other block of memory outside your 512 bytes, you'd want to point the stack to somewhere known. (It does start out valid, though, because interrupts are enabled.)
More importantly, you need to initialize DS to match your ORG setting, so ds:org reaches a linear address of 7C00. With org 0x7c00, that means you want DS=0. Otherwise instructions like mov [dx_L], bh would be using memory at some unknown location.
[BITS 16]
[ORG 7C00h]
[SEGMENT .text]
mov ax,0000h
mov ds,ax ; DS=0 to match ORG
mov ss,ax ; if you set SS:SP at all, do it back-to-back
mov sp,7C00h ; so an interrupt can't fire half way through.
; here do your stuff
l0:
hlt ; save power
jmp l0
Hope you are using VC or NC configured as IDE for NASM and not compiling/linking manually
This one is usable in MS-DOS so if you are running BOT SECTOR you out of luck. Still You can create a *.com executable debug and once its working in dos change to BOOT SECTOR...
see Is there a way to link object files for DOS from Linux? on how to setup MS-DOS Volkov commander to automatically compile and link your asm source code just by hitting enter on it ... You can also run it just by adding line to the vc.ext line ... but I prefer not to so you can inspect error log first
Convenient debugging
You can try to use MS-DOS (DOSBox) with ancient Borland Turbo C/C++ or Pascal and use their inline asm { .... } code which can be traced and stepped directly in the IDE. However it uses TASM (different syntax to NASM) and have some restrictions ...
Sadly I never saw any decent IDE for asm on x86 platform. The best IDE for asm I worked with was Herkules on ZX Spectrum ... was possible to done things even modern C++ IDEs doesnt have.
This is a part of a bootloader that I am studying from
`[ORG 0x00]
[BITS 16]
SECTION .text
jmp 0x07c0:START ; set CS(segment register) to 0x07C0 and jump to START label.
TOTALSECTORCOUNT:
dw 0x02
KERNEL32SECTORCOUNT:
dw 0x02
START:
mov ax, 0x07c0
mov ds, ax ; set DS(segment register) to the address of the bootloader.
mov ax, 0xb800
mov es, ax ; set ES(segment register) to the address of the video memory starting address.
; stack initialization
mov ax, 0x0000
mov ss, ax
mov sp, 0xfffe
mov bp, 0xfffe
; clear the screen
mov si, 0
CLEARSCREEN:
mov byte[es:si], 0
mov byte[es:si + 1], 0x0a
add si, 2
cmp si, 80 * 25 * 2
jl CLEARSCREEN
; print welcome message`
I don't understand the beginning: jmp 0x07C0:START How does it set the CS register?
And what are the two variables TOTALSECTORCOUNT and KERNEL32SECTORCOUNT for? They don't appear anywhere in the bootsector file and if I remove them, the bootloader fails to load the welcome message.
Removing the parts causes the OS to fail to load. So what is the significance of that jmp statement and the two variables?
``[ORG 0x00]
[BITS 16]
jmp START
START:
mov ax, 0x07c0
mov ds, ax ; set DS(segment register) to the address of the bootloader.
mov ax, 0xb800
mov es, ax ; set ES(segment register) to the address of the video memory starting address.
; stack initialization
mov ax, 0x0000
mov ss, ax
mov sp, 0xfffe
mov bp, 0xfffe
`
I am not great with assembly and usually use the AT&T syntax also. I have however written a bootloader before.
Hopefully you have learnt about the segmented addressing system used in 16 bit applications. The cs register holds the code segment. http://wiki.osdev.org/Segmentation
jmp 0x07C0:START ;This is a long jump
jmp segment:offset
A long jump sets the cs register to segment parameter and then does a jump to the offset parameter. When you do a short jump the cs register doesn't change. I assume that it would contain 0x0. You can use a short jump but you must tell your assembler or linker where the code will be run.
EDIT: After reading the code again there is the [org 0x00] line. This sets the cs register to 0x00 by default. If you wanted to use the short jump try changing this line to [org 0x7c00]
CS should already be set to 0x7c00 by the BIOS so the line:
jmp 0x07c0:START
can be replaced by:
jmp START
The two variables you mention must be use elsewhere in the code to load the kernel. However, it appears you haven't posted the whole code here.
Without seeing the rest of the bootsector code, we cannot help.
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I am writing a toy kernel for learning purposes, and I am having a bit of trouble with it. I have made a simple bootloader that loads a segment from a floppy disk (which is written in 32 bit code), then the bootloader enables the A20 gate and turns on protected mode. I can jump to the 32 bit code fine if I write it in assembler, but if I write it in C, I get a triple fault. When I disassemble the C code I can see that the first two instructions involve setting up a new stack frame. This is the key difference between the working ASM code and the failing C code.
I am using NASM v2.10.05 for the ASM code, and GCC from the DJGPP 4.72 collection for the C code.
This is the bootloader code:
org 7c00h
BITS 16
entry:
mov [drive], dl ;Save the current drive
cli
mov ax,cs ; Setup segment registers
mov ds,ax ; Make DS correct
mov ss,ax ; Make SS correct
mov bp,0fffeh
mov sp,0fffeh ;Setup a temporary stack
sti
;Set video mode to text
;===================
mov ah, 0
mov al, 3
int 10h
;===================
;Set current page to 0
;==================
mov ah, 5
mov al, 0
int 10h
;==================
;Load the sector
;=============
call load_image
;=============
;Clear interrupts
;=============
cli
;=============
;Disable NMIs
;============
in ax, 70h
and ax, 80h ;Set the high bit to 1
out 70h, ax
;============
;Enable A20:
;===========
mov ax, 02401h
int 15h
;===========
;Load the GDT
;===========
lgdt [gdt_pointer]
;===========
;Clear interrupts
;=============
cli
;=============
;Enter protected mode
;==================
mov eax, cr0
or eax, 1 ;Set the low bit to 1
mov cr0, eax
;==================
jmp 08h:clear_pipe ;Far jump to clear the instruction queue
;======================================================
load_image:
reset_drive:
mov ah, 00h
; DL contains *this* drive, given to us by the BIOS
int 13h
jc reset_drive
read_sectors:
mov ah, 02h
mov al, 01h
mov ch, 00h
mov cl, 02h
mov dh, 00h
; DL contains *this* drive, given to us by the BIOS
mov bx, 7E0h
mov es, bx
mov bx, 0
int 13h
jc read_sectors
ret
;======================================================
BITS 32 ;Protected mode now!
clear_pipe:
mov ax, 10h ; Save data segment identifier
mov ds, ax ; Move a valid data segment into the data segment register
mov es, ax
mov fs, ax
mov gs, ax
mov ss, ax ; Move a valid data segment into the stack segment register
mov esp, 90000h ; Move the stack pointer to 90000h
mov ebp, esp
jmp 08h:7E00h ;Jump to the kernel proper
;===============================================
;========== GLOBAL DESCRIPTOR TABLE ==========
;===============================================
gdt: ; Address for the GDT
gdt_null: ; Null Segment
dd 0
dd 0
gdt_code: ; Code segment, read/execute, nonconforming
dw 0FFFFh ; LIMIT, low 16 bits
dw 0 ; BASE, low 16 bits
db 0 ; BASE, middle 8 bits
db 10011010b ; ACCESS byte
db 11001111b ; GRANULARITY byte
db 0 ; BASE, low 8 bits
gdt_data: ; Data segment, read/write, expand down
dw 0FFFFh
dw 0
db 0
db 10010010b
db 11001111b
db 0
gdt_end: ; Used to calculate the size of the GDT
gdt_pointer: ; The GDT descriptor
dw gdt_end - gdt - 1 ; Limit (size)
dd gdt ; Address of the GDT
;===============================================
;===============================================
drive: db 00 ;A byte to store the current drive in
times 510-($-$$) db 00
db 055h
db 0AAh
And this is the kernel code:
void main()
{
asm("mov byte ptr [0x8000], 'T'");
asm("mov byte ptr [0x8001], 'e'");
asm("mov byte ptr [0x8002], 's'");
asm("mov byte ptr [0x8003], 't'");
}
The kernel simply inserts those four bytes into memory, which I can check as I am running the code in a VMPlayer virtual machine. If the bytes appear, then I know the code is working. If I write code in ASM that looks like this, then the program works:
org 7E00h
BITS 32
main:
mov byte [8000h], 'T'
mov byte [8001h], 'e'
mov byte [8002h], 's'
mov byte [8003h], 't'
hang:
jmp hang
The only differences are therefore the two stack operations I found in the disassembled C code, which are these:
push ebp
mov ebp, esp
Any help on this matter would be greatly appreciated. I figure I am missing something relatively minor, but crucial, here, as I know this sort of thing is possible to do.
Try using the technique here:
Is there a way to get gcc to output raw binary?
to produce a flat binary of the .text section from your object file.
I'd try moving the address of your protected mode stack to something else - say 0x80000 - which (according to this: http://wiki.osdev.org/Memory_Map_(x86)) is RAM which is guaranteed free for use (as opposed to the address you currently use - 0x90000 - which apparently may not be present - depending a bit on what you're using as a testing environment).
I'm one day into learning ASM and I've done a few tutorials, and even successfully modified the tutorial content to use jmp and cmp, etc instead of the MASM .if and .while macros.
I've decided to try and write something very, very simple to begin with before I continue with more advanced tutorials. I'm writing a Fibonacci number generator. Here is the source I have so far:
.386
.model flat, stdcall
option casemap :none
include \masm32\include\windows.inc
include \masm32\include\kernel32.inc
include \masm32\include\masm32.inc
includelib \masm32\lib\kernel32.lib
includelib \masm32\lib\masm32.lib
.code
start:
mov eax, 1
mov ecx, 1
_a:
push eax
add eax, ecx
pop ecx
; Jump to _b if there is an overflow on eax
; Print Values Here
jmp _a
_b:
push 0
call ExitProcess
end start
I intend to check for overflows on eax/ecx but right now I'm just interested in displaying the values of eax/ecx on the screen.
I know how to push the address of a constant string from .data and call StdOut which was the first example in the hello world tutorial, but this appears to be quite different (?).
There is this code provided by Microsoft itself
http://support.microsoft.com/kb/85068
Note that this code outputs AX register on 16 bit systems. But you can get the idea, you just need to convert AX value into ASCII characters by looping through each character. Skip the interrupts part and use your StdOut function.
mov dx, 4 ; Loop will print out 4 hex characters.
nexthex:
push dx ; Save the loop counter.
mov cl, 4 ; Rotate register 4 bits.
rol ax, cl
push ax ; Save current value in AX.
and al, 0Fh ; Mask off all but 4 lowest bits.
cmp al, 10 ; Check to see if digit is 0-9.
jl decimal ; Digit is 0-9.
add al, 7 ; Add 7 for Digits A-F.
decimal:
add al, 30h ; Add 30h to get ASCII character.
mov dl, al
;Use StdOut to print value of dl
;mov ah, 02h ; Prepare for interrupt.
;int 21h ; Do MS-DOS call to print out value.
pop ax ; Restore value to AX.
pop dx ; Restore the loop counter.
dec dx ; Decrement loop counter.
jnz nexthex ; Loop back if there is another character
; to print.
See here as well:
http://www.masm32.com/board/index.php?PHPSESSID=fa4590ba57dbaad4bc44088172af0b49&action=printpage;topic=14410.0
I have an assignment from my comp. system org. subject and unfortunately I'm kind of new when it comes to assembly language. I'm supposed to write a program that displays the numbers 0,2,4,6,8,10 respectively. How would I go about this?
Maybe this'll answer my question: (Reactions please)
.model small
.stack 100H
.data
.code
call proc
mov cx,5
mov dx,0
L1:
mov bx,2
add dx,bx
mov ah,02h
loop L1
int 21
endp
Go see your lecturer and/or tutor and ask for advice. That's what they're there for. You haven't given us anywhere near enough info to help you out.
Here's what I think your ABCD program should look like. I suggest you use it as a baseline to try to make a 0 2 4 ... version.
model proc
.stack 100H
.data
.call
main proc
mov cx,10 ; 10 loops only.
mov dx,40h ; start dx at 'A' - 1.
L1:
inc dx ; move to next character.
mov ah,02h ; int 21,02 is print character.
int 21h
loop L1 ; loop until cx is 0
mov ax,4c00h ; int 21,4c is exit with al holding exit code.
int 21
endp
When you've at least had a go at converting this, post the code and we'll critique what you've done.
If you're taught something, it never lasts but, if you learn something, it lasts forever (alcohol-addled braincells notwithstanding :-).
Int 21 is the DOS interrupt which allows assembler programs to use various DOS functions. It's conceptually a huge switch statement based on the AH register which is why you'll see things like Int 21 Fn 02, which means execute mov ah,2 followed by int 21.
Int 21 Fn 02 will take the contents of DL and output that to the screen. So the sequence:
mov ah,02h
mov dl,41h
int 21h
will output the 'A' character (0x41).
Similarly, Int 21 Fn 4c will exit the current running process.
I'm sure your class gave you some education here.
Can you code enough assembly to print one or two numbers?
Can you code enough to calculate the numbers, even if you can't print them?
Post that much code, and you may find help here.
Otherwise, you're asking others to actually do your homework for you.
Assembly language is a symbolic representation of the numeric machine codes and other constants needed to program a particular CPU (or architecture). So assembly language for Macs (most recently Intel's X86) is different from that used to on the iPhone - ARM.
Your teacher is also probably expecting you to realise the difference between the binary form of the number you will count with, and the ASCII format you will use to display to the screen.
You do know there is more than one flavor of "Assembly Language."
You can do it exactly like the program which prints A, B, C, D, etc.: except that instead of starting at 'A', start at '0; and instead of increasing by 1 each time (from 'A' to 'B'), increase by 2 (from '0' to '2').
After printing '0', '2', '4', '6', and '8', the next number that you want to print is '10'.
To print '10', you can print '1' followed by '0'. Or, instead of invoking int 21 with ah=2 (which prints one character at a time), you can set ah=9 to print a string (set ds:dx to a block of memory which contains "10$").
Later you suggested the following solution and asked for criticism:
.model small
.stack 100H
.data
.code
main proc
call defineuser1
call defineuser2
mov cx,5
userdefine1 proc
L1:
mov dx,0
mov bx,2
add dx,bx
mov ah,02h
loop L1
int 21h
endp
userdefine2 proc
mov ah, 4ch
int 21h
userdefine2
endp
My criticisms are as follows:
defineuser1 doesn't exist (I think you mean userdefine1)
setting cx needs to be inside (not before) the procedure
invoking int 21 needs to be inside (not outside) the loop
you need special handling for "10" as I mentioned above
There's a difference between '0' (the ASCII character/digit) and 0 (the number) ... you need to print the character/digit, not the number
You need to learn to test your code (write it, step through it with debugger, and debug it), preferably before you post questions about it.
You would have a counter beginning at zero and repeatedly increment it by two, printing the result.
.model small
.stack 100H
.code
.data
var2 DB "10$"
main proc
mov cx,4
mov ax,0
mov dl,al
add dl,30h
mov ah,02h
int 21h
mov ax,0
var1:
add ax,2
mov dl,al
add dl,30h
mov bx,ax
mov ah,2h
int 21h
mov ax,bx
loop var1
mov ax,#data
mov ds,ax
mov dx,offset var2
mov ah,09h
int 21h
main endp
end main
I'm new in computer science and when i saw this question i just wanted to try it. I have managed in doing it and here is the code
MOV AX, 0
MOV BX, 2
ADDLOOP:
ADD AX, BX
CMP AX, 10
JE DONE
JMP ADDLOOP
DONE:
Ok. That's my best attempt. Lots of details left out. I should also mention that I have no frigging clue how to print a char to the screen.
Like others have mentioned, you didn't specify which assembly language so I chose x86.
Finally, go talk to your instructors, they'll help you much more than we can.
Are you using a macro for the output?
should be something like...
mov eax, 0
myloop: cmp eax, 10
jg done
output macro eax
add eax, 2
jmp myloop
done:
of course that's for 8086 assembly.