這個完全是按照題目要求寫的cursor equ 45h

attrib equ 2fh

code segment

assume cs:code,ds:code

start:

jmp go

oldcur dw ?

old1c dw 2 dup(?)

newint1c:

pushf

call dword ptr cs:old1c

push ax

push bx

push cx

push dx

xor bh,bh

mov ah,3

int 10h

mov cs:oldcur,dx

mov ah,2

xor bh,bh

mov dx,cursor

int 10h

mov ah,2ch　　　　　

int 21h ；讀取系統時鐘

;-------------------------

;ch = hour

;cl = minute

;dh = second

;dl = 1/100 seconds

;------------------------

push dx

push cx;show hour

pop bx

push bx

call showhour

call showcolon;show minute

pop bx

call showother

call showcolon;show second

pop bx

call showother

mov dx,cs:oldcur

mov ah,2

xor bh,bh

int 10h

pop dx

pop cx

pop bx

pop ax

iretshowhour proc near

push bx

pop ax

shr ax,8

mov bl,10

div bl

;-------------------------

;ax = shang

;dx = yushu

;------------------------

add al,30h

call show

call curmove

mov ax,dx

add al,30h

call show

call curmove

ret

showhour endpshowother proc near

push bx

pop ax

shl ax,8

shr ax,8

mov bl,10

div bl

;-------------------------

;ax = shang

;dx = yushu

;------------------------

add al,30h

call show

call curmove

mov ax,dx

add al,30h

call show

call curmove

ret

showother endp

showcolon proc near

mov al,":"

call show

call curmove

ret

showcolon endpcurmove proc near

push ax

push bx

push cx

push dx

mov ah,3

mov bh,0

int 10h

inc dl

mov ah,2

int 10h

pop dx

pop cx

pop bx

pop ax

ret

curmove endpshow proc near

push ax

push bx

push cx

mov ah,09h

mov bx,attrib

mov cx,1

int 10h

pop cx

pop bx

pop ax

ret

show endpgo:

push cs

pop ds

mov ax,351ch ;取中斷向量

int 21h

mov old1c,bx　　　　　　；儲存原中斷向量

mov bx,es

mov old1c+2,bx

mov dx,offset newint1c　；置新的中斷向量

mov ax,251ch

int 21h

mov dx,offset go

sub dx,offset start

mov cl,4

shr dx,cl

add dx,11h

mov ax,3100h　　　　　　　；結束並駐留

int 21h

code ends

end start