总算把实验十四搞定了,嘿嘿
ASSUME CS:CODECODE SEGMENT
S DB 9,8,7,4,2,0
START:
MOV AX,CS
MOV DS,AX
MOV SI,OFFSET S
MOV DI,160*12+40*2
MOV CX,3
S1:
PUSH CX
MOV AX,
OUT 70H,AL
IN AL,71H
MOV AH,AL
MOV CL,4
SHR AH,CL
AND AL,00001111B
ADD AH,30H
ADD AL,30H
MOV BX,0B800H
MOV ES,BX
MOV BYTE PTR ES:,AH
MOV BYTE PTR ES:,2
MOV BYTE PTR ES:,AL
MOV BYTE PTR ES:,2
POP CX
CMP CX,1
JE S2
MOV BYTE PTR ES:,'/'
MOV BYTE PTR ES:,2
S2: INC SI
ADD DI,6
LOOP S1
MOV BYTE PTR ES:,' '
MOV CX,3
S3:
PUSH CX
MOV AX,
OUT 70H,AL
IN AL,71H
MOV AH,AL
MOV CL,4
SHR AH,CL
AND AL,00001111B
ADD AH,30H
ADD AL,30H
MOV BX,0B800H
MOV ES,BX
MOV BYTE PTR ES:,AH
MOV BYTE PTR ES:,2
MOV BYTE PTR ES:,AL
MOV BYTE PTR ES:,2
POP CX
CMP CX,1
JE OK
MOV BYTE PTR ES:,':'
MOV BYTE PTR ES:,2
OK:
NOP
INC SI
ADD DI,6
LOOP S3
MOV AX,4C00H
INT 21H
CODE ENDS
END START
希望大家晒晒自己的程序,我的程序有点愚笨,希望有更简洁版的版本 本帖最后由 梦想飞扬 于 2012-3-1 18:47 编辑
定义两个段db '00/00/00 00:00:00'
db 9,8,7,4,2,0
先把时间数据存储到前一个段中,再复制到显存 这位兄弟,你的这个方法好是好,但是你不觉得这样多做了一步吗?(多做了复制这一步)不如直接将时间写入显存,这样不是更好吗?你看? ASSUME CS:CODE
CODE SEGMENT
S DB 9,8,7,4,2,0
S0 DB '00/00/00 00:00:00'
START:
MOV AX,CS
MOV DS,AX
MOV SI,OFFSET S
MOV DI,OFFSET S0
MOV CX,3
S1:
PUSH CX
MOV AL,DS:
MOV AH,0
OUT 70H,AL
IN AL,71H
MOV AH,AL
MOV CL,4
SHR AH,CL
AND AL,00001111B
ADD AH,30H
ADD AL,30H
MOV DS:,AH
MOV DS:,AL
ADD DI,3
INC SI
POP CX
LOOP S1
MOV CX,3
S2:
PUSH CX
MOV AL,DS:
MOV AH,0
OUT 70H,AL
IN AL,71H
MOV AH,AL
MOV CL,4
SHR AH,CL
AND AL,00001111B
ADD AH,30H
ADD AL,30H
MOV DS:,AH
MOV DS:,AL
ADD DI,3
INC SI
POP CX
LOOP S2
MOV BX,0B800H
MOV ES,BX
MOV DI,160*12+40*2
MOV SI,OFFSET S0
MOV CX,17
S3:
MOV BYTE PTR AL,DS:
MOV BYTE PTR ES:,AL
MOV BYTE PTR ES:,2
INC SI
ADD DI,2
LOOP S3
MOV AX,4C00H
INT 21H
CODE ENDS
END START
按照‘梦想飞扬’的想法,我实现了一下,还是比较简单的,我之前那个程序写的好像有点烦 本帖最后由 梦想飞扬 于 2012-3-2 15:58 编辑
发下我自己做的吧
assume cs:code
code segment
d0: db '00/00/00 00:00:00'
d1: db 9,8,7,4,2,0
start: mov ax,cs
mov ds,ax
mov si,offset d0
mov di,offset d1
mov cx,6
push si
s:push cx
mov al,
out 70h,al ;写入数据
in al,71h ;读取数据
mov ah,al
mov cl,4
shr ah,cl
and al,00001111b
add ah,30h
add al,30h
mov byte ptr ,ah
mov byte ptr ,al
inc di
add si,3
pop cx
loop s
mov ax,0b800h
mov es,ax
mov bx,0
pop si
mov cx,17
s0:mov al,
mov es:,al
inc si
add bx,2
loop s0
ok:mov ax,4c00h
int 21h
code ends
end start
只是路过过来看看 学习了,不错的程序!{:1_1:} 大家的代码都有这么长。
我想,既然不可避免长的话,就简单明了算了。
如下:assume cs:code
code segment
start:
mov ax,0b800h
mov es,ax
mov si,160*12+81-17
mov al,9
out 70h,al
in al,71h
mov ah,al
mov cl,4
shr ah,cl
and al,00001111b
add ah,30h
add al,30h
mov byte ptr es:,ah
mov byte ptr es:2,al
mov byte ptr es:4,'/'
mov al,8
out 70h,al
in al,71h
mov ah,al
mov cl,4
shr ah,cl
and al,00001111b
add ah,30h
add al,30h
mov byte ptr es:6,ah
mov byte ptr es:8,al
mov byte ptr es:10,'/'
mov al,7
out 70h,al
in al,71h
mov ah,al
mov cl,4
shr ah,cl
and al,00001111b
add ah,30h
add al,30h
mov byte ptr es:12,ah
mov byte ptr es:14,al
mov byte ptr es:16,'/'
mov byte ptr es:18,' '
mov al,4
out 70h,al
in al,71h
mov ah,al
mov cl,4
shr ah,cl
and al,00001111b
add ah,30h
add al,30h
mov byte ptr es:20,ah
mov byte ptr es:22,al
mov byte ptr es:24,':'
mov al,2
out 70h,al
in al,71h
mov ah,al
mov cl,4
shr ah,cl
and al,00001111b
add ah,30h
add al,30h
mov byte ptr es:26,ah
mov byte ptr es:28,al
mov byte ptr es:30,':'
mov al,0
out 70h,al
in al,71h
mov ah,al
mov cl,4
shr ah,cl
and al,00001111b
add ah,30h
add al,30h
mov byte ptr es:32,ah
mov byte ptr es:34,al
mov ax,4c00h
int 21h
code ends
end start
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