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rkatsiri



Joined: 08 Feb 2016
Posts: 4
rkatsiri
Hi, I get a run error when i try running this program.
Is there a wrong command or a logic error in this code?
I'm new to assembly, so sorry if my question seems dum...

I wrote it in attempt to do this:
Write a program that takes the number n as input. Then it prints all the numbers x below n that have exactly 2 different integral divisors (Besides 1 and x).

Here's the code i wrote:

Code:
start:
    
call    read_hex        ; eax <- n
cmp eax,6h      
jb exit_program ;(if n is below 6, which is the smallest number with only 2 divisors)
        
mov     esi,2h          ; esi <- 2
dec     eax                     ; x <- n-- (x is the biggest number below n)

check_integers:
        
        mov ebx, eax            ;ebx <- x
        mov edi, 0h                     ; 0 divisiors
        div esi                 ; eax <- x/2 , edx <- remainder
        call print_eax  ; print just to check the code
        mov ecx, eax    ; ecx <- x/2
        mov eax, ebx    ; eax <- x
        
count_divisors:
        div ecx         ; eax <- x / (x/2) , edx <- remainder 
        mov eax, ebx    ; eax <- x
        cmp edx, 0h
        jnz skip_inc_divisors  ;(if remainder not equal 0)
        inc edi ; divisiors++

skip_inc_divisors:
        dec ecx         ; ecx--
        cmp ecx, 2h
        jnb count_divisors ;(if ecx not below 2)

        cmp edi, 2h;
        jne skip_print
        call print_eax  ; print x (print x if equals 2)

skip_print:
        dec eax         ; x--
        cmp eax,6h      
        jnb check_integers      ;(if eax not below 6, which is the smallest number with only 2 divisors)

exit_program:
    ; Exit the process:
        push    0
        call    [ExitProcess]

include 'training.inc'
    


Description:
Download
Filename: 0_just2divisors.asm
Filesize: 1.66 KB
Downloaded: 121 Time(s)

Post 09 Feb 2016, 12:46
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AsmGuru62



Joined: 28 Jan 2004
Posts: 1411
Location: Toronto, Canada
AsmGuru62
DIV instruction requires EDX to be set to zero, so 64 bits of whatever you dividing will fit into EDX:EAX pair.
I do not see you are doing it.
Can this be an issue?
Post 09 Feb 2016, 15:43
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rkatsiri



Joined: 08 Feb 2016
Posts: 4
rkatsiri
AsmGuru62, I've just added the following line in the beginning of the code:

Code:
mov edx, 0h
    


But unfortunately, i still get a run error,
right after the following output:

3
55555557[/code][/b]
Post 09 Feb 2016, 19:01
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AsmGuru62



Joined: 28 Jan 2004
Posts: 1411
Location: Toronto, Canada
AsmGuru62
Once DIV is complete - EDX may be set to a remainder - may be not zero.
So, you need to reset it to 0 before every DIV - you have another DIV below in the code.
Let me check the code in debugger.

Update:
I added two lines (marked with AG62) and code seems to run OK:
Code:
        ...

check_integers: 

        mov ebx, eax            ;ebx <- x
        mov edi, 0h                     ; 0 divisiors

        mov edx, 0h     ; AG62 - need EDX = 0 before DIV

        div esi                 ; eax <- x/2 , edx <- remainder
        call print_eax  ; print just to check the code
        mov ecx, eax    ; ecx <- x/2
        mov eax, ebx    ; eax <- x

count_divisors: 
        mov edx, 0h     ; AG62 - need EDX = 0 before DIV

        div ecx         ; eax <- x / (x/2) , edx <- remainder  
        mov eax, ebx    ; eax <- x 
        cmp edx, 0h 
        jnz skip_inc_divisors  ;(if remainder not equal 0) 
        inc edi ; divisiors++ 

        ...
    
Post 10 Feb 2016, 04:34
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Mikl___



Joined: 30 Dec 2014
Posts: 116
Mikl___
Code:
mov edx,eax     ; AG62 - need EDX = 0 before DIV
and edx,1 ; edx <- remainder
shr eax,1 ; eax <- x/2 
call print_eax  ; print just to check the code    
Post 10 Feb 2016, 06:00
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