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[OS programtabstrip显示器属性

Description: 关于显示输出 用 mov ah,02h int 21h 怎样才能输出大于10的数?请看一下我的程序,如果我输入十个a,就会出错。-on display output with mov ah, Check int 21 hours how to output more than 10 Look at the number of my procedure, if I enter a 10. will mistakes.
Platform: | Size: 1439 | Author: lp | Hits:

[assembly languagebaogeascii

Description: 一、用表格形式显示字符 1、题目:用表格形式显示ASCII字符 2、实验要求: 按15行*16列的表格形式显示ASCII码为10H-100H之间的所有字符,即以行为主的顺序用ASCII码递增的次序依次显示对应的字符.每16个字符为一行,每行中的相邻两个字符之间用空白符(ASCII为0)隔开. 3、提示: (1)显示每个字符可能使用功能号为02的显示输出功能调用,使用方法如下: MOV AH,02H MOV DL,输出字符的ASCII码 INT 21H 本题中可把DL初始化为10H,然后不断使其加1(用INC指令)以取得下一个字符的ASCII码. (2)显示空白符时,用其ASCII码为0置入DL寄存器.每行结束时,用显示回车和换行符结束本行并开始下一行. (3)由于逐个显示相继为ASCII字符时,需要保存并不断修改DL寄存器的内容,而显示空白、回车、换行符时也需要使用DL寄存器,为此可用堆栈来保存相继的ASCII字符。 具体用法是:在显示空白或回车、换行符前用指令PUSH DX把DL的内容保存到堆栈中去。在显示空白或回车、换行符后用指令恢复DL寄存器的原始内容。-a tabular form with a character show, entitled : tabular form showed two ASCII characters, experimental requirements : * 15 by 16 firms in tabular form for the show ASCII des-100H between all characters, that is, to act with the main sequence of ASCII increase in the order corresponding to the sequence of characters. Each of the 16 characters, and his party Each line of the adjacent between two characters with an empty site (ASCII 0) separated. 3, Tip : (1), showing each character may be used for 02 functions, the output function, use the following method : MOV AH, 02H MOV DL, Output ASCII characters INT 21H this issue can be put to initialize the DL 10 H, and then it increases one (using INC. Directive) to obtain under an ASCII characters. (2), showing a blank website , with its ASCII 0 p
Platform: | Size: 1454 | Author: 庄宁文 | Hits:

[Menu controlsum

Description: code segment assume cs:code sum_p proc near jcxz exit add ax,cx dec cx call sum_p exit: ret sum_p endp start: mov ax,2000h mov ds,ax mov ax,0 mov si,0 mov cx,7 call sum_p mov [si],ax mov ah,4ch int 21h code ends end start
Platform: | Size: 2278 | Author: 陨石 | Hits:

[assembly languagesystem

Description: 系统时间显示程序 利用BIOS的INT 1AH的2号系统功能调用,将计算机系统的时间参数(BCD码)送入寄存器。其中CH 和CL中保存的是小时数和分钟数;DH中保存的是秒钟数,将BCD码表示的时,分,秒转换成ASC I I码并送入屏幕显示。并用INT 21H 的7号功能调用判断输入回车并清屏
Platform: | Size: 41452 | Author: dong | Hits:

[WEB Code汇编语言课程设计1

Description: 程序首先定义一个hello的函数调用int 21h mov ah ,09来显示字符串,ds:dx定义字符串位置 来使用一个80×25的界面使整个程序更加美观 游戏主程序调用BIOS int 10 的9号功能实现对目标文本的颜色和定位,。游戏主程序可以用int 16 mov ah,00从键盘读取输入的字符在调用int 21 mov ah,01 让键盘输入显示在屏幕中 al=输入的字符,用cmp指令对输入的文本和目标文本进行比较,再利用选择语句将错误的输入字符显示为红色,将正确的输入字符显示为绿色。在整个字符串结束后直接退到dos环境-procedures hello first definition of a function call int 21 hours mov ah, 09 to show string, ds : dx definition string position to use a 80 x 25 interface to make the whole process more attractive game int main program called BIOS on the 9th of 10 functions to achieve the goal of text color and positioning. Game int main program can use 16 mov ah, 00 input from the keyboard to read characters in the call int 21 mov ah, 01 for keyboard input the screen were al = input characters using cmp directive of input text and the text more goals and used the wrong choice of words to the imported characters shown as red, to the correct input characters shown as green. The entire string after the end of direct retreated dos environment
Platform: | Size: 11952 | Author: 高赈寰 | Hits:

[assembly languageshujujiai

Description: 数据加密程序 p macro buf lea dx,buf mov ah,9 int 21h endm data segment pp db 0dh,0ah, welcome you to run this programme!$ qq db 0dh,0ah, when you input number,press enter.$ qw db 0dh,0ah, 6eh,52h,94h,52h,$ qa db 0dh,0ah, $ info1 db 0dh,0ah,0dh, input number:$ info2 db 0dh,0ah, output number:$ bufa db 20
Platform: | Size: 2895 | Author: ty | Hits:

[assembly language闰年计算

Description:

 data segment    ;定义数据段

    infon db 0dh,0ah,'Please input a year: $'
    Y db 0dh,0ah,'This is a leap year! $'
    N db 0dh,0ah,'This is not a leap year! $'
    w dw 0
    buf db 8
        db ?
        db 8 dup(?)
data ends
 
stack segment stack
    db 200 dup(0)
stack ends
 
code segment
          assume ds:data,ss:stack,cs:code
    start:mov ax,data
          mov ds,ax
         
          lea dx,infon ;在屏幕上显示提示信息
          mov ah,9
          int 21h
 
          lea dx,buf    ;从键盘输入年份字符串
          mov ah,10
          int 21h
         
          mov cl, [buf+1]
          lea di,buf+2
          call datacate
          call ifyears
          jc a1
         
          lea dx,n
          mov ah,9
          int 21h
          jmp exit
    a1:   lea dx,y
          mov ah,9
          int 21h
    exit: mov ah,4ch
          int 21h
      
 
 
 datacate proc near;
          push cx;                                                      
          dec cx
          lea si,buf+2
     tt1: inc si
          loop tt1
          ;lea si,cx[di]
          pop cx
    
     
          mov dh,30h
          mov bl,10
          mov ax,1
      l1: push ax
          sub byte ptr [si],dh
          mul byte ptr [si]
          add w,ax
          pop ax
          mul bl
          dec si
          loop l1
          ret
 datacate endp
 
 
   ifyears proc near
           push bx
           push cx
           push dx
           mov ax,w
           mov   cx,ax
           mov dx,0
           mov bx,4
           div bx
           cmp dx,0
           jnz lab1
           mov   ax,cx
           mov bx,100
           div bx
           cmp dx,0
           jnz lab2
           mov ax,cx
           mov bx,400
           div bx
           cmp dx,0
           jz lab2
     lab1: clc
           jmp lab3
     lab2: stc
     lab3: pop dx
           pop cx
           pop bx
           ret
   ifyears endp
code ends
   end start
     




[广告] 2008年最受关注楼盘


Platform: | Size: 10752 | Author: jackdawson0@ | Hits:

[Documents汇编语言课程设计1

Description: 程序首先定义一个hello的函数调用int 21h mov ah ,09来显示字符串,ds:dx定义字符串位置 来使用一个80×25的界面使整个程序更加美观 游戏主程序调用BIOS int 10 的9号功能实现对目标文本的颜色和定位,。游戏主程序可以用int 16 mov ah,00从键盘读取输入的字符在调用int 21 mov ah,01 让键盘输入显示在屏幕中 al=输入的字符,用cmp指令对输入的文本和目标文本进行比较,再利用选择语句将错误的输入字符显示为红色,将正确的输入字符显示为绿色。在整个字符串结束后直接退到dos环境-procedures hello first definition of a function call int 21 hours mov ah, 09 to show string, ds : dx definition string position to use a 80 x 25 interface to make the whole process more attractive game int main program called BIOS on the 9th of 10 functions to achieve the goal of text color and positioning. Game int main program can use 16 mov ah, 00 input from the keyboard to read characters in the call int 21 mov ah, 01 for keyboard input the screen were al = input characters using cmp directive of input text and the text more goals and used the wrong choice of words to the imported characters shown as red, to the correct input characters shown as green. The entire string after the end of direct retreated dos environment
Platform: | Size: 11264 | Author: 高赈寰 | Hits:

[OS programtabstrip显示器属性

Description: 关于显示输出 用 mov ah,02h int 21h 怎样才能输出大于10的数?请看一下我的程序,如果我输入十个a,就会出错。-on display output with mov ah, Check int 21 hours how to output more than 10 Look at the number of my procedure, if I enter a 10. will mistakes.
Platform: | Size: 1024 | Author: lp | Hits:

[assembly languagebaogeascii

Description: 一、用表格形式显示字符 1、题目:用表格形式显示ASCII字符 2、实验要求: 按15行*16列的表格形式显示ASCII码为10H-100H之间的所有字符,即以行为主的顺序用ASCII码递增的次序依次显示对应的字符.每16个字符为一行,每行中的相邻两个字符之间用空白符(ASCII为0)隔开. 3、提示: (1)显示每个字符可能使用功能号为02的显示输出功能调用,使用方法如下: MOV AH,02H MOV DL,输出字符的ASCII码 INT 21H 本题中可把DL初始化为10H,然后不断使其加1(用INC指令)以取得下一个字符的ASCII码. (2)显示空白符时,用其ASCII码为0置入DL寄存器.每行结束时,用显示回车和换行符结束本行并开始下一行. (3)由于逐个显示相继为ASCII字符时,需要保存并不断修改DL寄存器的内容,而显示空白、回车、换行符时也需要使用DL寄存器,为此可用堆栈来保存相继的ASCII字符。 具体用法是:在显示空白或回车、换行符前用指令PUSH DX把DL的内容保存到堆栈中去。在显示空白或回车、换行符后用指令恢复DL寄存器的原始内容。-a tabular form with a character show, entitled : tabular form showed two ASCII characters, experimental requirements :* 15 by 16 firms in tabular form for the show ASCII des-100H between all characters, that is, to act with the main sequence of ASCII increase in the order corresponding to the sequence of characters. Each of the 16 characters, and his party Each line of the adjacent between two characters with an empty site (ASCII 0) separated. 3, Tip : (1), showing each character may be used for 02 functions, the output function, use the following method : MOV AH, 02H MOV DL, Output ASCII characters INT 21H this issue can be put to initialize the DL 10 H, and then it increases one (using INC. Directive) to obtain under an ASCII characters. (2), showing a blank website , with its ASCII 0 p
Platform: | Size: 1024 | Author: 庄宁文 | Hits:

[Menu controlsum

Description: code segment assume cs:code sum_p proc near jcxz exit add ax,cx dec cx call sum_p exit: ret sum_p endp start: mov ax,2000h mov ds,ax mov ax,0 mov si,0 mov cx,7 call sum_p mov [si],ax mov ah,4ch int 21h code ends end start-code segment assume cs: codesum_p proc near jcxz exit add ax, cx dec cx call sum_pexit: retsum_p endpstart: mov ax, 2000h mov ds, ax mov ax, 0 mov si, 0 mov cx, 7 call sum_p mov [si], ax mov ah, 4ch int 21hcode ends end start
Platform: | Size: 2048 | Author: 陨石 | Hits:

[assembly languagesystem

Description: 系统时间显示程序 利用BIOS的INT 1AH的2号系统功能调用,将计算机系统的时间参数(BCD码)送入寄存器。其中CH 和CL中保存的是小时数和分钟数;DH中保存的是秒钟数,将BCD码表示的时,分,秒转换成ASC I I码并送入屏幕显示。并用INT 21H 的7号功能调用判断输入回车并清屏-err
Platform: | Size: 40960 | Author: dong | Hits:

[assembly languageshujujiai

Description: 数据加密程序 p macro buf lea dx,buf mov ah,9 int 21h endm data segment pp db 0dh,0ah, welcome you to run this programme!$ qq db 0dh,0ah, when you input number,press enter.$ qw db 0dh,0ah, 6eh,52h,94h,52h,$ qa db 0dh,0ah, $ info1 db 0dh,0ah,0dh, input number:$ info2 db 0dh,0ah, output number:$ bufa db 20 -Data encryption program p macro buflea dx, bufmov ah, 9int 21hendmdata segmentpp db 0dh, 0ah, welcome you to run this programme! $ Qq db 0dh, 0ah, when you input number, press enter. $ Qw db 0dh, 0ah, 6eh, 52h, 94h, 52h, $ qa db 0dh, 0ah, $ info1 db 0dh, 0ah, 0dh, input number: $ info2 db 0dh, 0ah, output number: $ bufa db 20
Platform: | Size: 3072 | Author: ty | Hits:

[Othershowtime

Description: 用汇编语言写的显示系统时间的小程序. 用于参考int 21h 的2ch号调用.-Using assembly language to write the time of the Display System applet. Used in reference to the int 21h call No. 2ch.
Platform: | Size: 1024 | Author: kandy | Hits:

[OS programzxc

Description: 小写字母变大写字母(汇编语言实现) (2008-05-14 21:57:52) 标签:杂谈 DATA SEGMENT PMT1 DB INPUT Small letter : ,0AH,0DH, $ STR1 DB 40H,0,40H DUP(0) PMT2 DB 0AH,0DH, Display capital letter : ,0AH,0DH STR2 DB 40H DUP(0), $ DATA ENDS STACK1 SEGMENT PARA STACK DW 20h DUP(0) STACK1 ENDS CODE SEGMENT ASSUME CS:CODE,DS:DATA START: MOV DX,DATA MOV DS,DX MOV DX, OFFSET PMT1 MOV AH,09H 提示输入小写字符 INT 21H MOV DX,OFFSET STR1 MOV AH,0AH 输入小写字符 INT 21H MOV CL,STR1+1 MOV CH,0 LEA SI,STR1+2 LEA DI,STR2 LOP:MOV AL,[SI] SUB AL,20H MOV [DI],AL INC SI INC DI LOOP LOP MOV DX,OFFSET PMT2 MOV AH,09H 输出大写字母 INT 21H MOV AH,4CH INT 21H CODE ENDS END START-Lowercase letters variable capital letters (compiled language) (2008-05-14 21:57:52) Tags: Talking DATA SEGMENT PMT1 DB INPUT Small letter:, 0AH, 0DH, $ STR1 DB 40H, 0,40 H DUP (0 ) PMT2 DB 0AH, 0DH, Display capital letter:, 0AH, 0DH STR2 DB 40H DUP (0), $ DATA ENDSSTACK1 SEGMENT PARA STACK DW 20h DUP (0) STACK1 ENDSCODE SEGMENT ASSUME CS: CODE, DS: DATASTART: MOV DX, DATA MOV DS, DX MOV DX, OFFSET PMT1 MOV AH, 09H prompted to enter lowercase characters INT 21H MOV DX, OFFSET STR1 MOV AH, 0AH enter lowercase characters INT 21H MOV CL, STR1+ 1 MOV CH, 0 LEA SI, STR1+ 2 LEA DI, STR2 LOP: MOV AL, [SI] SUB AL, 20H MOV [DI], AL INC SI INC DI LOOP LOP MOV DX, OFFSET PMT2 MOV AH, 09H output capital letters INT 21H MOV AH, 4CH INT 21HCODE ENDSEND START
Platform: | Size: 1024 | Author: 缺打打 | Hits:

[Communicationsfs

Description: DW 256 DUP(?) STACK1 ENDS DDATA SEGMENT MES1 DB The least number is:$ MES2 DB 0AH,0DH, The largest number is:$ NUMB DB 0D9H,07H,8BH,0C5H,0EBH,04H,9DH,0F9H DDATA ENDS CODE SEGMENT ASSUME CS:CODE,DS:DDATA START: MOV AX,DDATA MOV DS,AX MOV SI,OFFSET NUMB MOV CX,0008H JCXZ A4 MOV BH,[SI] MOV BL,BH A1: LODSB AL=DS:[SI],SI=SI+1 CMP AL,BH JBE A2 MOV BH,AL JMP A3 A2: CMP AL,BL JAE A3 MOV BL,AL A3: LOOP A1 A4: MOV DX,OFFSET MES1 show mes1 MOV AH,09H INT 21H MOV AL,BL show the least number AND AL,0F0H get the highest 4 bits SHR AL,4 CMP AL,0AH JB C2 ADD AL,07H C2: ADD AL,30H MOV DL,AL show character MOV AH,02H INT 21H MOV AL,BL AND AL,0FH get the lowest 4 bits CMP AL,0AH JB C3 ADD AL,07H C3: ADD AL,30H MOV DL,AL show character MOV AH,02H INT 21H MOV DX,OFFSET MES2 show mes2 MOV AH,09H INT 21H -DW 256 DUP (?) STACK1 ENDSDDATA SEGMENTMES1 DB The least number is: $ MES2 DB 0AH, 0DH, The largest number is: $ NUMB DB 0D9H, 07H, 8BH, 0C5H, 0EBH, 04H, 9DH, 0F9HDDATA ENDSCODE SEGMENT ASSUME CS: CODE, DS: DDATASTART: MOV AX, DDATA MOV DS, AX MOV SI, OFFSET NUMB MOV CX, 0008H JCXZ A4 MOV BH, [SI] MOV BL, BHA1: LODSB AL = DS: [SI], SI = SI+ 1 CMP AL, BH JBE A2 MOV BH, AL JMP A3A2: CMP AL, BL JAE A3 MOV BL, ALA3: LOOP A1A4: MOV DX, OFFSET MES1 show mes1 MOV AH, 09H INT 21H MOV AL, BL show the least number AND AL , 0F0H get the highest 4 bits SHR AL, 4 CMP AL, 0AH JB C2 ADD AL, 07H C2: ADD AL, 30H MOV DL, AL show character MOV AH, 02H INT 21H MOV AL, BL AND AL, 0FH get the lowest 4 bits CMP AL, 0AH JB C3 ADD AL, 07HC3: ADD AL, 30H MOV DL, AL show character MOV AH, 02H INT 21H MOV DX, OFFSET MES2 show mes2 MOV AH, 09H INT 21H
Platform: | Size: 1024 | Author: 张于 | Hits:

[assembly languageinterupte

Description: 汇编程序的中断程序的设计方法 int 21H的1号、2号、9号功能的使用 使用21H中断的9号功能,输出提示信息,并使用1号功能从键盘输入2个小于10的数,计算其和,并将和存放在存储单元R中。-Assembler interrupt process design method int 21H 1, 2, 9, features the use of 21H interruption of function on the 9th, the output message, and use the 1st function from the keyboard input of the two is less than 10 the number of calculated and will be stored in the memory cell and R in.
Platform: | Size: 4096 | Author: | Hits:

[Software Engineeringint21h

Description: 提供汇编INT功能所有描述,INT 21H里AH值全部都有,希望对大家有用,第一次上传-INT function to provide a compilation of all description, INT 21H value in AH, all of the above, I hope useful for all of us, the first upload
Platform: | Size: 10240 | Author: 周超 | Hits:

[SCM110316

Description: 简单16位汇编,通过int 21h的中断调用,进行文件读取,并写入到其他文件。-Simple 16-bit assembly, through the int 21h interrupt call, the file read and write to other files.
Platform: | Size: 1024 | Author: q | Hits:

[assembly languageint-21h--ah=9

Description: 汇编小程序 INT 21H 的9号功能输出,应用了BUFFER-Compilation of small program 9 function output
Platform: | Size: 1024 | Author: MA | Hits:
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