CODEWITHSUNDEEP
loopsassemblydivisionx86-16
Justo Guillermo montoya
Justo Guillermo montoya

Reputation: 49

Executing the DIV instruction in 8086 assembly

My program prints the msg3 statement (PutStr msg3), but does not proceed to the 

DIV CX

instruction in my program.
Is there something I'm doing incorrectly with that register?
Or should the instruction be 

DIV [CX]

instead or do I not have the compare and jump conditions set correctly?

prime_loop:
        sub AX,AX       ;clears the reg to allow the next index of the array
        sub CX,CX      ;clears counter to decrement starting from number of the value for array
        mov AX, [test_marks+ESI*4]      ;copy value of array at index ESI into reg
        mov CX, [test_marks+ESI*4]     ;copy value of array at index ESI into reg for purposes of counting down
    check_prime:
        dec CX 
        nwln
        PutStr msg3
        div WORD CX   ;divide value of EAX by ECX
        cmp DX,0     ;IF the remainder is zero
        je chck_divisor    ;check to see divisor 'ECX'
        sub AX,AX   ;else clear quotient register EAX
        sub DX,DX    ;clear remainder register
       mov AX,[test_marks+ESI*4]     ;move the number of the current iteration back into EAX
        jmp check_prime    ;start again from loop
    chck_divisor:
        cmp CX,1
        jne prime_loop     ;if the divisor is not 1 then it is not a prime number so continue with iterations
        PutInt AX     ;else print the prime_num
        PutStr
        inc ESI
        jmp prime_loop
    done:
        .EXIT

Upvotes: 2

Views: 1503

Answers (2)

Sep Roland
Sep Roland

Reputation: 39306

These are some points about your code:

  • If this is indeed 8086 assembly then instructions like mov AX, [test_marks+ESI*4] that use scaled indexed addressing simply don't exist!

  • The scale by 4 suggests that your array is filled with doublewords, yet you use just a word. This could be what you want, but it looks suspicious.

  • Let's hope no array element is 1 because if so, then the div cx instruction will trigger an exception (#DE). Because you don't test the CX register for becoming 0.

  • In the check_prime loop only the 1st iteration lacks the zeroing of DX in order to give a correct quotient.

The solution will depend on the targetted architecture 8086 or x86. Now your program is a mix of both!

Upvotes: 1

rcgldr
rcgldr

Reputation: 28921

It's possible that since DX is not zeroed before div, that you're getting overflow. I don't know how your environment handles overflow.

Upvotes: 0

Related Questions