Question

Complete the following Intel assembly language program

which determines whether the byte sized operand stored

in memory location 'number' is prime or not. The program

will write the value of 0 into the memory location 'answer'

if the number is not prime, otherwise the initial value of '1' will

be left unmodified (indicating that the number is prime). The program

makes use of the DIV instruction to determine the value of quotient

and remainder when dividing the number by 2,3,4,... (number -1) .

section .data

; put your data in this section using

; db , dw, dd directions

number db 5

answer db 1 ; 1 means number is prime, 0 means number is not prime

section .bss

; put UNINITIALIZED data here using

section .text

global _start

_start:

mov esi, number ; get the offset of number into esi  

keith: mov eax, 0 ; clear the entire eax register

mov al, [esi] ; get the number from memory into al

mov dl, al ; put it inside dl as well

mov bl, 2 ; bl holds each divisor starting from 2

loopy: div bl ; ax / bl with quot in al and rem in ah

and ax, 1111111100000000b ; isolate the rem in ah with a AND mask

; to determine whether the remainder is 0

; YOU COMPLETE THE REST OF THE CODE

HINT: The program is to implement the following high-level

pseudocode:

prime = true ;

divisor = 2 ;

while ( divisor < number )

{

if ( the remainder of number / divisor is 0 )

{

prime = false ; // we found a divisor which evenly divides number

} // so therefore number cannot be prime

   

divisor = divisor + 1 ; // check the next divisor and keep looping

}

Answer 1

.... jz non_prime
.... inc bl
.... cmp bl, dl
.... jl loopy
[add code here to report a prime number]
non_prime:
[add code here to report a non-prime number]


Both the assembly code and the C pseudocode are pretty n00b-like.

1. There is no need to clear EAX if you are going to be doing all 8 and 16-bit arithmetic.

2. There's no point in AND AX, 0FF00h, when AND AH,AH does the same thing with a shorter instruction

3. There is no point in the pseudocode staying in the loop when a number is found to be non-prime. Assembly language is all goto-based, so you don't have to pretend that goto is evil and "avoid a break; statement".

4.One leading test for divisibility by 2 is the only even divisor you need to test. If a number isn't divisible by 2, it isn't going to have any other even divisors, either.

5. You can really take advantage of the fact that the DIV instruction gives you a quotient as well as a remainder. If the quotient is (strictly) less than the divisor, that means the divisor is greater than the square root of the original number. If you haven't seen a divisor yet, you aren't going to. You can stop after sqrt(n) tries and it doesn't cost a single machine cycle or a register to compare against.

Those last two This cuts the complexity for testing a prime number p down from p-2 divisions to approximately sqrt(p)/2 divisions. So, to find that 241 is prime will take 8 divisions instead of 239. I don't count division by 2 because that can be done simply by testing bit 0 of the number.

if you find my answer correct please give this a tumbs up, or we can discuss more about the answer.Thank You