The algorithm is not sieve (which would create 2^31 locations and strike off multiples of primes). Rather this is a space optimized algorithm that repeatedly divides each number n from 2 to sqrt(n) to determine if a number is prime. This algorithm uses no additional storage than the 165 bytes to hold the code.
.model tiny
.code
.386
org 100h
start:
main proc
xor eax,eax
mov al,3
@@m2:
call isprime
jnc @@m1
call printdword
call printcrlf
@@m1:
inc eax
jnz @@m2
mov ax,4c00h
int 21h
main endp
isprime proc
pushad
mov esi,eax ; save number
xor ecx,ecx
mov ebx,ecx
mov edx,ecx ; edx = 0
call sqrt
mov cx,ax ; save sqrt
mov bl,2 ; ebx = 2 (divisor)
@@i2:
mov eax,esi ; restore number
cmp ecx,ebx ; is sqrt < divisor?
jl @@i4 ; yes, no need to divide any more
div ebx
or edx,edx
jz @@i1
xor edx,edx
inc ebx
jnz @@i2
@@i4:
stc
jmp @@i3
@@i1:
clc
@@i3:
popad
ret
isprime endp
sqrt proc
push ebx
push ecx
xor ecx,ecx
mov ebx,ecx
@@s2:
inc ebx
sub eax,ebx
jl @@s1
inc cx
inc ebx
jmp @@s2
@@s1:
mov ax,cx
pop ecx
pop ebx
ret
sqrt endp
printdword proc
pushad
xor ebx,ebx
mov cx,bx
mov bl,10
@@p1:
xor edx,edx
div ebx
push dx
inc cx
or eax,eax
jnz @@p1
@@p2:
pop ax
add al,'0'
mov dl,al
mov ah,02
int 21h
loop @@p2
popad
ret
printdword endp
printcrlf proc
pusha
mov dl,13
mov ah,02
int 21h
mov dl,10
mov ah,2
int 21h
popa
ret
printcrlf endp
end start
