I'm trying to solve one simple task in assembly (TASM), namely:
There is a natural number with a range in the word, determine the sum of digits in the second degree by this number.
I want to output to DOS the result of adding 6^2 + 1^2 + 3^2 + 1^2 + 3^2.
The below code can only output a number in DOS, no more, given by our instructor.
;stack segments
stk segment stack
db 128 dup(?)
stk ends
;data segment
data segment para public 'data'
x dw 61313
DThousands dw ?
Thousands dw ?
Hundreds dw ?
Decades dw ?
Units dw ?
result dw ?
data ends
;command segment
code segment para public 'code'
assume cs:code, ds:data, ss:stk
begin:
mov ax, data
mov ds, ax
mov ax, x ; заносим число x в регистр ax
mov result, ax ; заносим в зарезервированный участок памяти result значение из ax
mov ax, result ; меняем значение
xor cx, cx ;MOV CX, 0
mov bx, 10 ; bx = 10
m_do_while:
xor dx, dx ; обнуление dx
div bx ; деление ax на bx
push dx ; заталкиваем dx в стек
inc cx ; увеличиваем cx на 1
cmp ax, 0 ; сравниваем регистр ax с нулем
jne m_do_while ; выполняем условный переход
mov ah, 2 ; помещаем в регистр ah 2
m_for:
pop dx ; достаем из стека значение dx
add dx, 30h ; прибавляем к dx 30h
int 21h ; системное прерывание
loop m_for ; цикл
back:
;end of program
mov ax, 4C00h
int 21h
code ends
end begin
You're already getting the digits one at a time with that printing loop. Addition is associative so it doesn't matter what order you get them in, you can add starting with the least-significant digit.
digit_sum:
mov ax, x ; input in AX
mov bx, 10 ; base 10
xor cx, cx ; sum
.sumloop:
xor dx, dx
div bx ; quotient in AX, remainder (the digit) in DX
;; With 386
;imul dx, dx ; requires 386
;add cx, dx ; sum += digit^2
;; Without 386
xchg ax, dx
mul al ; result in AX. DX untouched. single-digit numbers fit in AL
add cx, ax ; sum += digit^2
mov ax, dx
test ax, ax
jne .sumloop
;;; sum in CX
ret
Then print cx efficiently, e.g. by converting into a buffer starting from the end and then making one print system call. (How do I print an integer in Assembly Level Programming without printf from the c library?). I wouldn't recommend that clunky push/pop 2-loop method you show in the question, but it's popular and does work. Anyway, mov ax, cx would put the sum in AX.
You could even get some code-reuse by using a divide-by-10-and-push loop like you have, or like in Assembly 8086 | Sum of an array, printing multi-digit numbers. The first time, use it to get digits which you pop and square -> sum. The second time, use it to generate digits of the sum, which you pop and print. (But writing a function that leaves a variable amount of stuff on the stack is tricky; you could pop the return address at the start of the function, then push/ret. Or just make it a macro that you use twice, so it inlines both places.)
If you wanted to be 8086-compatible but tune for more recent Intel CPUs (where xchg is 3 uops and thus costs about the same as 3 mov instructions): Instead of xchg ax,dx you could use mov si, ax / mov ax, dx, then mul/add, then mov ax, si. For actual ancient 8086, xchg is great: smaller is faster (except for really slow instructions like mul and div) and xchg-with-ax is only 1 byte.
Of course if you actually care about speed you'd use a multiplicative inverse to divide by 10. And for actual 8086 where mul is quite slow (but not as slow as div), you might use a lookup-table of squares to save a mul in that part:
; given a digit in DX, add its square to CX, indexing a table of words
mov si, dx
shl si,1
add cx, [table + si]
Or with just a table of bytes, trading 1 byte of extra code-size for a smaller table and 1 fewer byte of data loaded (break even on 8088, except for prefetch differences):
mov si, dx
add cl, [table + si]
adc ch, 0 ; carry to the high half of CX