CODEWITHSUNDEEP
assemblyx86
Amit Singh Tomar
Amit Singh Tomar

Reputation: 8610

Trouble with prime number program in Assembly

I am learning assembly language on x86 and made a simple programe that will tell number is prime or not.

I think I am doing right but still not getting the desired result,below is code

section .bss
b db
section .data
x db "Number is Prime",10,0
y db "Number is not Prime",10,0
z db "value is=%d",10,0

section .text
global main
extern printf
main:
mov eax,17
mov ebx,2
loop:
mov [b],eax
div ebx
mov eax,[b]
cmp edx,0
jz Print_not_Prime
inc ebx

 cmp ebx,17
 jnz loop
 push x
 call printf
 add esp,4
 ret

 Print_not_Prime:
 push y
 call printf
 add esp,4
 ret

In above code I am checking with number 17 and output of programe telling its not prime number.

Could anybody let me know where I am doing wrong

Upvotes: 0

Views: 2532

Answers (2)

Sergey L.
Sergey L.

Reputation: 22552

You need to zero edx before each divide. The dividend for div comes as a double-register operand in edx:eax.

If you have a remainder in there from the last trial division then it will screw up your results.

Also it's better to store your dividend in a register (ecx, esi, edi are still unused) or at least on the stack then in memory which as Michael pointed out is insufficient to store a dword.

Upvotes: 2

Michael
Michael

Reputation: 58467

I see at least two problems with this:

b db
....
mov [b],eax

You're only reserving space for a byte at b but storing a dword (4 bytes). You should use dd instead of db.

div ebx

You should use cdq (or xor edx,edx) prior to div in order to clear edx, since this division will divide edx:eax by ebx.

Upvotes: 2

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