Programming Forums > Application Development > Assembly

akioshin · Oct 26th, 2007, 11:32 PM

iv been trying to code this but everytime i run the program it wont sort or it wont swap, can someone at least give me an example with sorting 5 integers using array, so that i will use this as my reference for my project,my project is that to sort names in alpabetical order,

DaWei · Oct 26th, 2007, 11:59 PM

Well, we can't tell what is wrong with your program, unless we're mind readers, because you haven't provided any code for us to look at.

No, we won't write your code for you. We will help you with what you write. You have joined our community and asked for free help. I perceive that you have not read our rules. That seems rather rude, does it not?

akioshin · Oct 27th, 2007, 12:22 AM

oh im realy sory heres my code for sorting 5 integers using array, its not finish yet im still testing for the first and second index to swap

(Toggle Plain Text)

extrn readnum:near
extrn writenum:near
extrn new_line:near
extrn clearscreen:near
public num
.286
.model small

printStr macro str
   pusha
   lea dx, str
   mov ah, 9
   int 21h
   popa
endm

.stack 0
.data

	msg db, 10, 13, 'Please enter one digit number: $'
	num dw 0
	ar1 dw 5 dup(?)

.code


	mov ax, @data
	mov ds, ax


	mov bx, offset ar1
	mov si, offset ar1
	mov cx, 5

aloop:

	printStr msg
	call readnum
	mov ax, num
	mov [bx], ax
	add bx, 2
	loop aloop

check:
	
	add bp, 2
	mov ax, [bx]	
	cmp ax, [si]
	jg swap

swap:

	mov ax, [bx]
	push ax
	mov ax, [bx+2]
	mov [bx], ax
	pop ax
	mov [bx+2], ax
	

	jmp disp1



;for displaying the array
disp1:
	mov bx, offset ar1
	mov cx, 5
disp:
	
	mov ax, word ptr [bx]
	mov num, ax
	call new_line
	call writenum	
	add bx, 2
	loop disp


	mov ah, 4ch
	int 21h
end

DaWei · Oct 27th, 2007, 12:32 AM

Notice here:

(Toggle Plain Text)

check:
	
	add bp, 2
	mov ax, [bx]	
	cmp ax, [si]
	jg swap

swap:

You always wind up at swap, regardless of the relationship between the two items tested. I think you've also lost a handle on where bx points. I think you've also lost view of what happens after a swap.

Have you tried stepping through this with a debugger? I highly recommend it.

akioshin · Oct 27th, 2007, 2:08 AM

ah ya im just a newbie at this, can you check this one this is my second attempt and can you advise me of wat gud books or site for assembly

(Toggle Plain Text)

extrn readnum:near
extrn writenum:near
extrn new_line:near
extrn clearscreen:near
public num
.286
.model small

printStr macro str
   pusha
   lea dx, str
   mov ah, 9
   int 21h
   popa
endm

.stack 0
.data

	msg db, 10, 13, 'Please enter one digit number: $'
	num dw 0
	ar1 dw 5 dup(?)

.code


	mov ax, @data
	mov ds, ax


	mov bx, offset ar1
	sub si, si
	sub di, di
	mov cx, 5
aloop:

	printStr msg
	call readnum
	mov ax, num
	mov [bx], ax
	add bx, 2
	loop aloop
	
	mov di, 2

comp:
	mov ah, [bx+si]
	mov al, [bx+di]
	cmp ah, al

	jge swap
swap:

	
	mov [bx+si], al
	mov [bx+di], ah
	jmp disp1

	

;for displaying the array
disp1:
	mov bx, offset ar1
	mov cx, 5
disp:
	
	mov ax,[bx]
	mov num, ax
	call new_line
	call writenum	
	add bx, 2
	loop disp


	mov ah, 4ch
	int 21h
end

RobEasy · Oct 27th, 2007, 3:48 AM

http://www.xs4all.nl/~smit/asm01001.htm

Honestly, you are a lot farther along than the tutorial is but more info can never hurt.

DaWei · Oct 27th, 2007, 5:06 AM

You're missing the point.

(Toggle Plain Text)

comp:
	mov ah, [bx+si]
	mov al, [bx+di]
	cmp ah, al

	jge swap  If it >= we jump to swap, else we fall through to swap.
swap:               What have you accomplished?

	
	mov [bx+si], al
	mov [bx+di], ah
	jmp disp1   Here you jump where?  Right where you'd go if you removed this

	

;for displaying the array
disp1:

You need to mentally trace your code, asking yourself what happens at every step. Conditionals are the heart of processing. If they don't result in different paths, they're worthless.

lectricpharaoh · Oct 28th, 2007, 8:39 AM

Re-read DaWei's post. Then try plotting out the logical flow before you write any further code. I'll give you an example using a bubble sort, since it's very easy to understand. It should look something like this:

(Toggle Plain Text)

set unsorted items = array size
while (unsorted items > 1)
  current item index = 0
  while (current item index < unsorted items)
    if (array[current item index] > array[current item index+1])
      swap (array[current item index], array[current item index+1])
    increment current item index
  decrement unsorted items

Here is the process translated to assembly (assumes all segment registers are correctly initialized):

(Toggle Plain Text)

sort:
  mov cx, [arraysize] ; load size of array
  cmp cx, 2           ; and ensure we have two or more elements
  jl  done            ; we're done if there are fewer than two elements
  mov bx, [array]     ; point at start of array- note 'array' is a pointer,
                      ; so we will dereference it in the inner loop below
nextpass:
  mov si, 0           ; SI is our current item index
nextelement:
  mov ax, [bx+si]     ; fetch current element
  cmp ax, [bx+si+1]   ; and compare to following element
  jge noswap          ; if the second is not smaller, skip the swap
  xchg ax, [bx+si+1]
  mov ax, [bx+si]
noswap:
  inc si              ; increment index for next element
  cmp cx, si          ; and check if we've finished the current pass
  jl nextelement      ; still on this pass, so process next element
  loop nextpass       ; do next pass if any remain, else we're done
done:

Be warned. First, it's been ages since I've played with assembly, so there may very well be errors. Second, I've used NASM syntax, but it's easy to translate to TASM syntax. Just remember that if it's in square brackets, it refers to a memory location, and if it's not, it's an assembly-time constant, ie an offset in TASM parlance (even labels are offsets, albeit into the code segment). I'll leave the rest (printing, etc) up to you; I'm not going to do everything for you.

As a final suggestion, I'd really advise you to comment your assembly code. It's not like a higher-level language where you can derive a lot of meaning from variable names and other constructs; poorly commented asm is damn near opaque when you come back to it later.

DaWei · Oct 28th, 2007, 9:36 AM

My guess is he's trying to accomplish an insertion sort.

lectricpharaoh · Oct 28th, 2007, 9:43 AM

Quote:

Originally Posted by DaWei

My guess is he's trying to accomplish an insertion sort.

Actually, I was kinda hoping that, since the array was so small (perhaps below the magical cut-off point), bubble sort would actually be more efficient.

That's my story, and I'm sticking to it. It's got nothing to do with me being a lazy ass, and not wanting to write something marginally more complex in assembly.

Oct 26th, 2007, 11:32 PM	#1
akioshin Newbie Join Date: Oct 2007 Posts: 6 Rep Power: 0	tasm sorting 5 integers with array iv been trying to code this but everytime i run the program it wont sort or it wont swap, can someone at least give me an example with sorting 5 integers using array, so that i will use this as my reference for my project,my project is that to sort names in alpabetical order,

Oct 26th, 2007, 11:59 PM	#2
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Re: tasm sorting 5 integers with array Well, we can't tell what is wrong with your program, unless we're mind readers, because you haven't provided any code for us to look at. No, we won't write your code for you. We will help you with what you write. You have joined our community and asked for free help. I perceive that you have not read our rules. That seems rather rude, does it not? __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Oct 27th, 2007, 12:22 AM	#3
akioshin Newbie Join Date: Oct 2007 Posts: 6 Rep Power: 0	Re: tasm sorting 5 integers with array oh im realy sory heres my code for sorting 5 integers using array, its not finish yet im still testing for the first and second index to swap (Toggle Plain Text) extrn readnum:near extrn writenum:near extrn new_line:near extrn clearscreen:near public num .286 .model small printStr macro str pusha lea dx, str mov ah, 9 int 21h popa endm .stack 0 .data msg db, 10, 13, 'Please enter one digit number: $' num dw 0 ar1 dw 5 dup(?) .code mov ax, @data mov ds, ax mov bx, offset ar1 mov si, offset ar1 mov cx, 5 aloop: printStr msg call readnum mov ax, num mov [bx], ax add bx, 2 loop aloop check: add bp, 2 mov ax, [bx] cmp ax, [si] jg swap swap: mov ax, [bx] push ax mov ax, [bx+2] mov [bx], ax pop ax mov [bx+2], ax jmp disp1 ;for displaying the array disp1: mov bx, offset ar1 mov cx, 5 disp: mov ax, word ptr [bx] mov num, ax call new_line call writenum add bx, 2 loop disp mov ah, 4ch int 21h end extrn readnum:near extrn writenum:near extrn new_line:near extrn clearscreen:near public num .286 .model small printStr macro str pusha lea dx, str mov ah, 9 int 21h popa endm .stack 0 .data msg db, 10, 13, 'Please enter one digit number: $' num dw 0 ar1 dw 5 dup(?) .code mov ax, @data mov ds, ax mov bx, offset ar1 mov si, offset ar1 mov cx, 5 aloop: printStr msg call readnum mov ax, num mov [bx], ax add bx, 2 loop aloop check: add bp, 2 mov ax, [bx] cmp ax, [si] jg swap swap: mov ax, [bx] push ax mov ax, [bx+2] mov [bx], ax pop ax mov [bx+2], ax jmp disp1 ;for displaying the array disp1: mov bx, offset ar1 mov cx, 5 disp: mov ax, word ptr [bx] mov num, ax call new_line call writenum add bx, 2 loop disp mov ah, 4ch int 21h end Last edited by DaWei; Oct 27th, 2007 at 12:23 AM. Reason: Added code tags. Read the rules again.

Oct 27th, 2007, 12:32 AM	#4
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Re: tasm sorting 5 integers with array Notice here: (Toggle Plain Text) check: add bp, 2 mov ax, [bx] cmp ax, [si] jg swap swap: check: add bp, 2 mov ax, [bx] cmp ax, [si] jg swap swap: You always wind up at swap, regardless of the relationship between the two items tested. I think you've also lost a handle on where bx points. I think you've also lost view of what happens after a swap. Have you tried stepping through this with a debugger? I highly recommend it. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Oct 27th, 2007, 2:08 AM	#5
akioshin Newbie Join Date: Oct 2007 Posts: 6 Rep Power: 0	Re: tasm sorting 5 integers with array ah ya im just a newbie at this, can you check this one this is my second attempt and can you advise me of wat gud books or site for assembly (Toggle Plain Text) extrn readnum:near extrn writenum:near extrn new_line:near extrn clearscreen:near public num .286 .model small printStr macro str pusha lea dx, str mov ah, 9 int 21h popa endm .stack 0 .data msg db, 10, 13, 'Please enter one digit number: $' num dw 0 ar1 dw 5 dup(?) .code mov ax, @data mov ds, ax mov bx, offset ar1 sub si, si sub di, di mov cx, 5 aloop: printStr msg call readnum mov ax, num mov [bx], ax add bx, 2 loop aloop mov di, 2 comp: mov ah, [bx+si] mov al, [bx+di] cmp ah, al jge swap swap: mov [bx+si], al mov [bx+di], ah jmp disp1 ;for displaying the array disp1: mov bx, offset ar1 mov cx, 5 disp: mov ax,[bx] mov num, ax call new_line call writenum add bx, 2 loop disp mov ah, 4ch int 21h end extrn readnum:near extrn writenum:near extrn new_line:near extrn clearscreen:near public num .286 .model small printStr macro str pusha lea dx, str mov ah, 9 int 21h popa endm .stack 0 .data msg db, 10, 13, 'Please enter one digit number: $' num dw 0 ar1 dw 5 dup(?) .code mov ax, @data mov ds, ax mov bx, offset ar1 sub si, si sub di, di mov cx, 5 aloop: printStr msg call readnum mov ax, num mov [bx], ax add bx, 2 loop aloop mov di, 2 comp: mov ah, [bx+si] mov al, [bx+di] cmp ah, al jge swap swap: mov [bx+si], al mov [bx+di], ah jmp disp1 ;for displaying the array disp1: mov bx, offset ar1 mov cx, 5 disp: mov ax,[bx] mov num, ax call new_line call writenum add bx, 2 loop disp mov ah, 4ch int 21h end Last edited by DaWei; Oct 27th, 2007 at 2:10 AM. Reason: Added code tags. Try to get the point.

Oct 27th, 2007, 3:48 AM	#6
RobEasy Programmer Join Date: May 2006 Location: The US duhhhhh! Posts: 42 Rep Power: 0	Re: tasm sorting 5 integers with array http://www.xs4all.nl/~smit/asm01001.htm Honestly, you are a lot farther along than the tutorial is but more info can never hurt. __________________ Work Hard... Play Harder! Last edited by RobEasy; Oct 27th, 2007 at 3:55 AM. Reason: I am completely stupid

Oct 27th, 2007, 5:06 AM	#7
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Re: tasm sorting 5 integers with array You're missing the point. (Toggle Plain Text) comp: mov ah, [bx+si] mov al, [bx+di] cmp ah, al jge swap If it >= we jump to swap, else we fall through to swap. swap: What have you accomplished? mov [bx+si], al mov [bx+di], ah jmp disp1 Here you jump where? Right where you'd go if you removed this ;for displaying the array disp1: comp: mov ah, [bx+si] mov al, [bx+di] cmp ah, al jge swap If it >= we jump to swap, else we fall through to swap. swap: What have you accomplished? mov [bx+si], al mov [bx+di], ah jmp disp1 Here you jump where? Right where you'd go if you removed this ;for displaying the array disp1: You need to mentally trace your code, asking yourself what happens at every step. Conditionals are the heart of processing. If they don't result in different paths, they're worthless. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Oct 28th, 2007, 8:39 AM	#8
lectricpharaoh Caffeinated Neural Net Join Date: Jun 2005 Location: Wet west coast of Canada Posts: 1,119 Rep Power: 5	Re: tasm sorting 5 integers with array Re-read DaWei's post. Then try plotting out the logical flow before you write any further code. I'll give you an example using a bubble sort, since it's very easy to understand. It should look something like this: (Toggle Plain Text) set unsorted items = array size while (unsorted items > 1) current item index = 0 while (current item index < unsorted items) if (array[current item index] > array[current item index+1]) swap (array[current item index], array[current item index+1]) increment current item index decrement unsorted items set unsorted items = array size while (unsorted items > 1) current item index = 0 while (current item index < unsorted items) if (array[current item index] > array[current item index+1]) swap (array[current item index], array[current item index+1]) increment current item index decrement unsorted items Here is the process translated to assembly (assumes all segment registers are correctly initialized): (Toggle Plain Text) sort: mov cx, [arraysize] ; load size of array cmp cx, 2 ; and ensure we have two or more elements jl done ; we're done if there are fewer than two elements mov bx, [array] ; point at start of array- note 'array' is a pointer, ; so we will dereference it in the inner loop below nextpass: mov si, 0 ; SI is our current item index nextelement: mov ax, [bx+si] ; fetch current element cmp ax, [bx+si+1] ; and compare to following element jge noswap ; if the second is not smaller, skip the swap xchg ax, [bx+si+1] mov ax, [bx+si] noswap: inc si ; increment index for next element cmp cx, si ; and check if we've finished the current pass jl nextelement ; still on this pass, so process next element loop nextpass ; do next pass if any remain, else we're done done: sort: mov cx, [arraysize] ; load size of array cmp cx, 2 ; and ensure we have two or more elements jl done ; we're done if there are fewer than two elements mov bx, [array] ; point at start of array- note 'array' is a pointer, ; so we will dereference it in the inner loop below nextpass: mov si, 0 ; SI is our current item index nextelement: mov ax, [bx+si] ; fetch current element cmp ax, [bx+si+1] ; and compare to following element jge noswap ; if the second is not smaller, skip the swap xchg ax, [bx+si+1] mov ax, [bx+si] noswap: inc si ; increment index for next element cmp cx, si ; and check if we've finished the current pass jl nextelement ; still on this pass, so process next element loop nextpass ; do next pass if any remain, else we're done done: Be warned. First, it's been ages since I've played with assembly, so there may very well be errors. Second, I've used NASM syntax, but it's easy to translate to TASM syntax. Just remember that if it's in square brackets, it refers to a memory location, and if it's not, it's an assembly-time constant, ie an offset in TASM parlance (even labels are offsets, albeit into the code segment). I'll leave the rest (printing, etc) up to you; I'm not going to do everything for you. As a final suggestion, I'd really advise you to comment your assembly code. It's not like a higher-level language where you can derive a lot of meaning from variable names and other constructs; poorly commented asm is damn near opaque when you come back to it later. __________________ And once again, Probability proves itself willing to sneak into a back alley and service Drama as would a copper-piece harlot. - Vaarsuvius, Order of the Stick

Oct 28th, 2007, 9:36 AM	#9
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Re: tasm sorting 5 integers with array My guess is he's trying to accomplish an insertion sort. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

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Programming Forums > Application Development > Assembly

tasm sorting 5 integers with array