[friiz1337]

I'm a beginner in Assembly and I'm trying to make my program execute the following formula using 3 already specified numbers for a,b and h: ((a+b)/2)*h
The result, I'm trying to store in a variable called result. However I keep getting wrong output (0004). I have commented the way I believe the code works. Could I get some help?
Here is my code:

Code

section .data
a: dw 3
b: dw 7
h: dw 4
result: dw 0

section .text
global main
main:
mov ecx, [a]  ;ecx = 3
mov edx, [b]  ;edx = 7
add ecx, edx  ;ecx + edx = 10
mov eax, ecx  ;eax = 10
mov edx, 0   ;edx = 0
mov ebx, eax  ;ebx = 10
mov eax, 2   ;eax = 2
div eax       ;ebx/eax = 10/2 = 5
mov eax, eax  ;eax = 5
mov ebx, [h]  ;ebx = 4
mul ebx     ;5x4 = 20
mov [result], ebx  ;result = 20
mov eax, 1
mov ebx, 0
int 80h

Thank you

[SIM-3364521]

with the [] you dereference the variable.
lets say you have something like:
a db 5
if you want to move the value of a in eax you would do:
mov eax, a
if you would do something like this:
mov eax, [a]
you would actually move the value at the address 0x5 in to the register.

That was the first thing i noticed i hope that was the only one and that you get it to work!
I would suggest you to have a look at the different addressing modes you can use in assembler:
https://www.tutorialspoint.com/assembly_..._modes.htm
(edit: just read through the tutorial on that site and its very basic but you should get the idea.
On the wikipedea page about assembler there is a more in depth explanation of different addressing modes also some that arent listed on tutorialspoint)

hope that was helpful

[friiz1337]

with the [] you dereference the variable.
lets say you have something like:
a db 5
if you want to move the value of a in eax you would do:
mov eax, a
if you would do something like this:
mov eax, [a]
you would actually move the value at the address 0x5 in to the register.

That was the first thing i noticed i hope that was the only one and that you get it to work!
I would suggest you to have a look at the different addressing modes you can use in assembler:
https://www.tutorialspoint.com/assembly_..._modes.htm
(edit: just read through the tutorial on that site and its very basic but you should get the idea.
On the wikipedea page about assembler there is a more in depth explanation of different addressing modes also some that arent listed on tutorialspoint)

hope that was helpful

I don't think it will work, I tried both ways, none of them got me anything. I thought the [] are to specify the value that is hold in the variable (abh in this case)

[HFDone]

your problem is the registers you using! read the description
https://www.aldeid.com/wiki/X86-assembly...ctions/div
https://www.aldeid.com/wiki/X86-assembly...ctions/mul

[friiz1337]

your problem is the registers you using! read the description
https://www.aldeid.com/wiki/X86-assembly...ctions/div
https://www.aldeid.com/wiki/X86-assembly...ctions/mul

Are you talking about multiplying and division only? Or adding too?
Thanks

---

I managed to edit the code a bit, I get a value(30) close to the intended one(20). However I can't seem to understand where miscalculation is happening.. Any ideas?

Code

section .data
a: dw 3
b: dw 7
h: dw 4
result: dw 0
divide: dw 2

section .text
global main
main:
mov ebx, [a]  ;ebx = 3
mov edx, [b]  ;edx = 7
add ebx, edx  ;ebx = 10
mov edx, 0    ;edx = 0
mov eax, ebx  ;eax = 10
mov ecx, [divide]    ;ecx = 2
div ecx       ;eax= 5 edx=0
mov edx, 0    ;edx = 0
mov eax, eax  ;eax =5
mul eax       ;result stored in edx after multiplication
mov [result], edx
int 80h

[HFDone]

a: dd 3
b: dd 7
h: dd 4
result: dd 0

Code

mov ebx, [a]  ;ecx = 3
mov edx, [b]  ;edx = 7
add ebx, edx  ;ebx = 10
mov edx, 0    ;edx = 0
mov eax, ebx  ;eax = 10
mov ecx, 2    ;ecx = 2
div ecx       ;eax = 5
mov edx, 0    ;edx = 0
mul eax       ;eax = 25
mov [result], eax

you dont use the h var for the mul
change
mov edx, 0    ;edx = 0
mul eax       ;eax = 25
to
mov edx, [h] ;edx = 4
mul edx ;eax = 20

[friiz1337]

a: dd 3
b: dd 7
h: dd 4
result: dd 0

Code

mov ebx, [a]  ;ecx = 3
mov edx, [b]  ;edx = 7
add ebx, edx  ;ebx = 10
mov edx, 0    ;edx = 0
mov eax, ebx  ;eax = 10
mov ecx, 2    ;ecx = 2
div ecx       ;eax = 5
mov edx, 0    ;edx = 0
mul eax       ;eax = 25
mov [result], eax

you dont use the h var for the mul

I understand, check my new code. The answer should be 20, not 25. I got to 30.
Thank you

[HFDone]

I understand, check my new code. The answer should be 20, not 25. I got to 30.
Thank you

i edit my reply

[friiz1337]

i edit my reply

Thank you dude.

---

i edit my reply

Is it possible to explain why do you used DD instead of DW?

[HFDone]

cuz you are using 32 bit (dword) registers and your vars are 16 bit (word)

[friiz1337]

cuz you are using 32 bit (dword) registers and your vars are 16 bit (word)

Thank you, I got it 🎉