10.1 顯示字符串
; multi-segment executable file template.
data segment
; add your data here!
db 'Welcome to masm!',0
ends
stack segment
dw 128 dup(0)
ends
code segment
start:
; set segment registers:
mov ax,stack
mov ss,ax
mov sp,256
mov ax,data
mov ds, ax
; add your code here
mov si,0
mov dh,8 ; line=8
mov dl,3 ; column=3
mov bl,2 ;color=g
call show_str
;end of your code
mov ax, 4c00h ; exit to operating system.
int 21h
;show string
;dh:line(0-24) dl:column(0,79),bl:color ds:si string start addr
;es:bx->0b800:offset
show_str:
;save registers
push ax
push es
push bx
push dx
;init es
mov ax,0b800h
mov es,ax
;calc display memory offset according to line and columm
mov al,dh
mov cl,160
mul cl; 160 bytes per line
mov dh,0
add ax,dx ;offset = 160*line + 2*column
add ax,dx
mov bx,ax
l:
;string end?
mov cl,[si]
mov ch,0
jcxz ok
;begin write
mov al,cl
mov es:[bx],al
;write color
mov es:[bx+1],cl
;update indexs
inc bx
inc bx
inc si
jmp short l
ok:
;recover registers
pop dx
pop bx
pop es
pop ax
ret
ends
end start ; set entry point and stop the assembler.
10.2 解決除法溢出問題
; multi-segment executable file template.
data segment
; add your data here!
pkey db "press any key...$"
ends
stack segment
dw 128 dup(0)
ends
code segment
start:
; set segment registers:
mov ax, data
mov ds, ax
mov ax,stack
mov ss,ax
mov sp,256
; add your code here
mov ax,4240h
mov dx,000fh
mov cx, 0ah
call divdw
mov ax, 4c00h ; exit to operating system.
int 21h
;被除數 dx:ax 除數 cx 商 dx：ax 餘數 cx
divdw:
;save
push bx
push si
;calc dx/cx
;save ax->bx
mov bx,ax
mov ax,dx
mov dx,0
div cx
;save 商->si
mov si,ax
;calc （餘數+低16位）除法
mov ax,bx
div cx
;
mov cx,dx
mov dx,si
pop si
pop bx
ret
ends
end start ; set entry point and stop the assembler.
10.3 數值顯示
這裏有個地方需要說明一下，就是計算的時候，先算出來的是數值的低位，而存儲的時候先存儲的必須是高位。這裏用壓棧出棧的方法巧妙地解決了這個問題。
data segment
; add your data here!
dw 12366,456,23456
db 30 dup (0)
ends
stack segment
dw 128 dup(0)
ends
code segment
start:
; set segment registers:
mov ax,stack
mov ss,ax
mov sp,256
mov ax,data
mov ds, ax
; add your code here
mov cx,3
mov di,0
mov si,6
l3:
mov dx,0 ; not supported 32 bit now.
mov ax,[di]
call itoa
mov dh,8 ; line=8
mov ax,di
add dh,al
mov dl,3 ; column=3
mov bl,7 ;color=white/black
call show_str
;update si
add si,bp
add dh,1
inc di
inc di
loop l3
;end of your code
mov ax, 4c00h ; exit to operating system.
int 21h
;convert int to string
;ax: int dx:0
;ds:si string begin add
;bp:strlen
itoa:
;save
push bx
push cx
push dx
push di
mov bp,0
mov bx,10
l1:
div bx
inc bp
mov cx,ax
push dx
mov dx,0;clear 被除數的高16位
jcxz itoa_ok
jmp short l1
itoa_ok:
mov cx,bp ;strlen
;begin pop into ds:si
mov di,si
l2:
pop ax
add ax,30h
mov [di],al
inc di
loop l2
mov al,0
mov [di],al ;last 0 added
pop di
pop dx
pop cx
pop bx
ret
;show string
;dh:line(0-24) dl:column(0,79),bl:color ds:si string start addr
;es:bx->0b800:offset
show_str:
;save registers
push cx
push ax
push es
push bx
push dx
push bx ;save color
;init es
mov ax,0b800h
mov es,ax
;calc display memory offset according to line and columm
mov al,dh
mov cl,160
mul cl; 160 bytes per line
mov dh,0
add ax,dx ;offset = 160*line + 2*column
add ax,dx
mov bx,ax
pop ax ;restore color
mov ah,al
l:
;string end?
mov cl,[si]
mov ch,0
jcxz ok
;begin write
mov al,cl
mov es:[bx],al
;write color
mov es:[bx+1],ah
;update indexs
inc bx
inc bx
inc si
jmp short l
ok:
;recover registers
pop dx
pop bx
pop es
pop ax
pop cx
ret
ends
end start ; set entry point and stop the assembler.