中国DOS联盟

都是我以前练习时自己编得玩的。
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dseg segment
cury db 0
curx db 0
luy db 10
lux db 30
rdy db 15
rdx db 40
cont db 5
dseg ends
clear macro
mov ah ,6
mov al ,0
mov ch ,0
mov cl,0
mov dh,24
mov dl,79
mov bh,7
int 10
endm
scroll macro
mov h,6
mov al,cont
mov ch,luy
mov cl,lux
mov dh,rdy
mov dl,rdx
mov bh,70h
int 10h
endm
postc macro
mov ah,2
mov dh,cury
mov dl,curx
mov bh,0
int 10
endm
;--------------------------------
cseg segment
assume cs:cseg,ds:dseg
main proc far
push ds
sub ax,ax
push ax
mov ax,dseg
mov ds,ax
clear
scroll cont
postc
input:
mov ah,0
int 16
cmp ah,abh
jnz no_left
cmp curx,0
jnz l1
cmp cury,0
jnz l3
jmp input
l1:
dec curx
jmp l5
l3:
dec cury
mov curx,79
l5:
postc
jmp input
no_left:
cmp ah,4dh
jnz no_right
cmp curx,79
jnz r1
cmp cury,24
jnz r3
jmp input
r1:
inc curx
jmp r5
r3:
inc cury
mov curx,0
r5:
postc
jmp input
no_right:
cmp ah,48h
jnz no_up
cmp cury,0
jz up1
dec cury
postc
up1:
jmp input
no_up:
cmp ah,50h
jnz no_down
cmp cury,24
jz d1
inc cury
postc
d1:
jmp input
no_down:
cmp ah,4fh
jz setxy
cmp ah,01
jnz disp
ret
disp:
mov ah,0ah
mov bh,0
mov cx,1
int 10h
inc curx
mov al,curx
cmp al,rdx
jl next
scroll 1
mov al,lux
mov curx,al
next:
postc
jmp input
setxy:
mov al,cury
mov bl,curx
cmp al,luy
jl s1
cmp bl,lux
jl s2
mov rdy,al
mov rdx,bl
jmp new
s1:
cmp bl,rdx
jle s3
jmp input
s2:
cmp al,rdy
jg s5
s3:
mov luy,al
mov lux,bl
new:
mov al,rdy
sub al,luy
inc al
mov cont ,al
clear
scroll cont
s5:
jmp input
main endp
;-----------------------------
cseg ends
end main
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事隔多年，我也忘光了，自己编译一下试试就知道了。

这个和上面的差不多，但可以肯定还是有不同的，编译后对比一下试试就知道了
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dseg segment
cury db 0
curx db 0
luy db 10
lux db 30
rdy db 15
rdx db 40
cont db 5
dseg ends
stack segment para stack'stack'
db 64 dup(?)
stack ends
clear macro
mov ah ,6
mov al ,18h
mov ch ,0
mov cl,0
mov dh,24
mov dl,79
mov bh,7
int 10h
endm
scroll macro cont
mov ah,6
mov al,cont
mov ch,luy
mov cl,lux
mov dh,rdy
mov dl,rdx
mov bh,00011111b
int 10h
endm
postc macro
mov ah,2
mov dh,cury
mov dl,curx
mov bh,0
int 10h
endm
;--------------------------------
cseg segment
assume cs:cseg,ds:dseg
main proc far
push ds
sub ax,ax
push ax
mov ax,dseg
mov ds,ax
clear
scroll cont
postc
input:
mov ah,0
int 16h
cmp ah,4bh
jnz no_left
cmp curx,0
jnz l1
cmp cury,0
jnz l3
jmp input
l1:
dec curx
jmp l5
l3:
dec cury
mov curx,79
l5:
postc
jmp input
no_left:
cmp ah,4dh
jnz no_right
cmp curx,79
jnz r1
cmp cury,24
jnz r3
jmp input
r1:
inc curx
jmp r5
r3:
inc cury
mov curx,0
r5:
postc
jmp input
no_right:
cmp ah,48h
jnz no_up
cmp cury,0
jz up1
dec cury
postc
up1:
jmp input
no_up:
cmp ah,50h
jnz no_down
cmp cury,24
jz d1
inc cury
postc
d1:
jmp input
no_down:
cmp ah,4fh
jz sxy
cmp ah,01h
jz edd
cmp al,0dh
jnz eed
neew:
inc cury
mov ah,cury
cmp ah,rdy
jle no_change
scroll 1
dec cury
mov ah,cury
no_change:
dec lux
mov al,lux
mov curx,al
inc lux

eed: jmp disp

edd: ret
sxy: jmp setxy
disp:
mov ah,0ah
mov bh,0
mov cx,1
int 10h
inc curx
mov al,curx
cmp al,rdx
jng next
scroll 1
mov al,lux
mov curx,al

next:
postc
jmp input
setxy:
mov al,cury
mov bl,curx
cmp al,luy
jl s1
cmp bl,lux
jl s2
mov rdy,al
mov rdx,bl
jmp new
s1:
cmp bl,rdx
jle s3
jmp input
s2:
cmp al,rdy
jg s5
s3:
mov luy,al
mov lux,bl
new:
mov al,rdy
sub al,luy
inc al
mov cont ,al
clear
scroll cont
s5:
jmp input
main endp
;-----------------------------
cseg ends
end main

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大概浏览了一下代码
好像第一个是图形模式，第二个是文本模式
不过第一段程序
scroll macro
mov h,6
那块似乎有点问题，
是不是改成：
scroll macro cont
mov ah,6

我最近正好在准备汇编语言的学期考试，被BIOS中断那块搞得很是头疼
不知道kickout对这块熟不熟，不过看程序用到不少

多帮帮我哈，如果这回考试顺利过关的话我请客哦

我试试代码先