Assembly 8086 program that calculates c = ((22-7)/5)*113

hi i really have no idea how this is working

dat开发者_运维知识库a segment
 db 22
 db 7
 db 5
 db 113
data ends
code segment
 assume cs:code,ds:data
start:mov si,500H
 mov di,1500H
 mov ah,0
 mov al,22
 cbw
 sub ax,7
 mov cx,5
 idiv cx
 mov bx,113
 imul bx
 mov dl,ax
 mov [di],al
 code ends
end start

i really don't know what's going on in this program so any help would be appreciated

In general code not looks correct. But anyway let's review main parts:

data segment - declares 3 bytes of data that would be used for calc (but unfortunately doesn't used at all). db - is declaration of byte (compare with dw - d eclare w ord, ...) code segment - declares that program code starts there

totally for perform calculation you need only following:

mov AX, 22; place number 22 to processor register AX
sub AX, 7; now we have subtract 7 from AX and place it back to AX
mov CX, 5; 5 is placed to CX register
xor DX, DX; per comment of @GJ
idiv cx ; divide AX / CX
mov bx,113; BX now contains 113
imul bx; at last mul result of AX on BX and place back to AX

Here are some useful tips:

• mov dst, src copies the src into the dest
• operand *dst*, src performs the operand on the src and the dest and copies the result into the dest. For instance, sub ax, 7 calculates ax - 7 and puts the result in ax.
• idiv src multiplies the src by ax, and puts the result in ax. imul does the same with multiplication.

This should be enough to get you started.

Simple:

mov word[c],((22-7)/5)*113