. : : Assembly Language : : .  |  首页  |  我提出的问题  |  我参与的问题  |  我的收藏  |  消息中心   |  游客  登录  |

刷新 | 提问 | 未解决 | 已解决 | 精华区 | 搜索 |

《汇编语言》论坛 －＞外中断

管理员： assembly

[回复本贴]　[收藏本贴]　[管理本贴]　[关闭窗口]

主题 : :　 求解实验15  [已解决] 回复[ 3次 ]   点击[ 429次 ]   liuquanhao7232 [帖 主]   [ 发表时间：2010-08-15 17:49 ]   [引用]   [回复]   [ top ]　 荣誉值：8 信誉值：0 注册日期：2010-07-12 08:47 因为没有DOS系统，所以大家帮下忙验证一下。 assume cs:code,ss:stack stack segment     db 128 dup (0) stack ends code segment         start:mov ax,stack               mov ss,ax               mov sp,128               mov ax,code               mov ds,ax               mov si,offset int9a               mov ax,0               mov es,ax               mov di,204h               mov cx,offset int9aend-offset int9a               cld               rep movsb                              push es:[9*4]               pop es:[200]               push es:[9*4+2]               pop es:[202]                              cli               mov word ptr es:[9*4],204h               mov word ptr es:[9*4+2],0               sti                              mov ax,4c00h               int 21h                                       int9a:push ax               push bx               push cx               push es                              in al,60h                              pushf               call dword ptr cs:[200h]                              mov ax,40h               mov es,ax               mov al,es:[17]               cmp al,01000000b               jne int9aend                    ;确定是否是输入大写                              cmp al,9Eh               jne int9anext                  ;确定是否是放开A键                             s:mov cx,2000               mov ax,0b800h               mov es,ax               mov bx,1               mov byte ptr es:[bx],41H               add bx,2               loop s                           ;这个在整个屏幕显示A                         next:cmp al,1               jne int9aend               mov ax,0               mov es,ax               push es:[200]               pop es:[9*4]               push es:[202]               pop es:[9*4+2]                 ;输入Eac则还原int 9例程的地址                     int9aend:pop es               pop cx               pop bx               pop ax                              iret                code ends end start liuquanhao7232 [第1楼]   [ 回复时间：2010-08-15 17:51 ]   [引用]   [回复]   [ top ]　 荣誉值：8 信誉值：0 注册日期：2010-07-12 08:47 不好意思，改正一下。 assume cs:code,ss:stack stack segment     db 128 dup (0) stack ends code segment         start:mov ax,stack               mov ss,ax               mov sp,128               mov ax,code               mov ds,ax               mov si,offset int9a               mov ax,0               mov es,ax               mov di,204h               mov cx,offset int9aend-offset int9a               cld               rep movsb                              push es:[9*4]               pop es:[200]               push es:[9*4+2]               pop es:[202]                              cli               mov word ptr es:[9*4],204h               mov word ptr es:[9*4+2],0               sti                              mov ax,4c00h               int 21h                                       int9a:push ax               push bx               push cx               push es                              in al,60h                              pushf               call dword ptr cs:[200h]                              mov ax,40h               mov es,ax               mov al,es:[17]               cmp al,01000000b               jne int9aend                    ;确定是否是输入大写                              cmp al,9Eh               jne int9anext                  ;确定是否是放开A键                             s:mov cx,2000               mov ax,0b800h               mov es,ax               mov bx,1               mov byte ptr es:[bx],41H               add bx,2               loop s                           ;这个在整个屏幕显示A                     int9next:cmp al,1               jne int9aend               mov ax,0               mov es,ax               push es:[200]               pop es:[9*4]               push es:[202]               pop es:[9*4+2]                 ;输入Eac则还原int 9例程的地址                     int9aend:pop es               pop cx               pop bx               pop ax                              iret                code ends end start liuquanhao7232 [第2楼]   [ 回复时间：2010-08-15 17:53 ]   [引用]   [回复]   [ top ]　 荣誉值：8 信誉值：0 注册日期：2010-07-12 08:47 额。。这个正确的。在50行那个。是int9anext 不好意思，改正一下。 assume cs:code,ss:stack stack segment     db 128 dup (0) stack ends code segment         start:mov ax,stack               mov ss,ax               mov sp,128               mov ax,code               mov ds,ax               mov si,offset int9a               mov ax,0               mov es,ax               mov di,204h               mov cx,offset int9aend-offset int9a               cld               rep movsb                              push es:[9*4]               pop es:[200]               push es:[9*4+2]               pop es:[202]                              cli               mov word ptr es:[9*4],204h               mov word ptr es:[9*4+2],0               sti                              mov ax,4c00h               int 21h                                       int9a:push ax               push bx               push cx               push es                              in al,60h                              pushf               call dword ptr cs:[200h]                              mov ax,40h               mov es,ax               mov al,es:[17]               cmp al,01000000b               jne int9aend                    ;确定是否是输入大写                              cmp al,9Eh               jne int9anext                  ;确定是否是放开A键                             s:mov cx,2000               mov ax,0b800h               mov es,ax               mov bx,1               mov byte ptr es:[bx],41H               add bx,2               loop s                           ;这个在整个屏幕显示A                     int9next:cmp al,1               jne int9aend               mov ax,0               mov es,ax               push es:[200]               pop es:[9*4]               push es:[202]               pop es:[9*4+2]                 ;输入Eac则还原int 9例程的地址                     int9aend:pop es               pop cx               pop bx               pop ax                              iret                code ends end start liuquanhao7232 [第3楼]   [ 回复时间：2010-08-15 18:48 ]   [引用]   [回复]   [ top ]　 荣誉值：8 信誉值：0 注册日期：2010-07-12 08:47 此贴由 贴主 于 [ 2010-08-15 18:48 ] 结贴。 结贴原因：问题已解决 得分情况： 此问题已结贴！

	Copyright © 2006-2024   ASMEDU.NET  All Rights Reserved