计算数字的平方并将其存储在组件8086中的dw中

Question

I am trying to calculate the square of 5 using brute force, so I tried this but it does not store anything in dw and I define dw like this result dw 2 dup(?)

mov ax, 5
mov bx, 5
mov cx, 1
mov dx, ax
loop1:
 add ax, dx

 cmp bx, cx
 je endLoop

 add cx, 1
 jmp loop1

endLoop:
 mov result, ax

Answer 1

calculate the square of 5

5^2 = 25 but your code produces 30

Your loop does 5 additions in total (1x in the fallthrough and 4x in the jump back).
Since AX starts out with its original value, you get too much!

Either start AX at zero or do 1 iteration less:

 mov ax, 5
 mov bx, 5
 mov cx, 1
 mov dx, ax
 XOR AX, AX   ; start at zero
loop1:
 add ax, dx

 cmp bx, cx
 je endLoop

 add cx, 1
 jmp loop1

-

 mov ax, 5
 mov bx, 5
 mov cx, 1+1  ; one iteration less
 mov dx, ax
loop1:
 add ax, dx

 cmp bx, cx
 je endLoop

 add cx, 1
 jmp loop1

it does not store anything in dw and I define dw like this result dw 2 dup(?)

Make sure to setup DS with code like this:

mov ax, data    ; maybe you'll need 'mov ax, @data'
mov ds, ax

Try an alternative memory addressing using square brackets:

mov [result], ax

---

From comment:

i am trying to retrieve the result like this
lea si, result
inc si
mov dl, [si]
mov ah, 2h
int 21h

Why do you fetch the second byte of the result? That will probably be plain zero.
Moreover the result has 2 digits, therefore you need to ouput 2 characters.

This is a solution from one of my other recent answers:

mov ax, [result]     ; AX=25
mov bl, 10
div bl               ; AL=2   AH=5
add ax, "00"         ; AL='2' AH='5'
mov dx, ax
mov ah, 02h
int 21h              ; Outputs the character in DL='2'
mov dl, dh
mov ah, 02h
int 21h              ; Outputs the character in DL='5'