我回来了
这个easy
assume cs:codesg
datasg segment
db "Beginner's All-purpose Symbolic Instruction Code.",0
datasg ends
codesg segment
begin:
mov ax,datasg
mov ds,ax
mov si,0
call letterc
mov ax,4c00h
int 21h
letterc:
...


- [yugong] \\\|/// \\ .-.- // . 09/22 16:27
- [musicvs] 我也来和楼主庆祝~ \\\|/// \\ .-. 09/21 23:54
- [genius023] 呵呵 谢谢happy~~ 再贴一个完成dword型数据转化到字符型的,方便取用 assume c 09/21 07:38
- [musicvs] 我也来鼓励一下楼主:不错! 嘻^_^ 09/20 22:02
- [游客] 很好 09/20 09:58
- [genius023] 这个也很好 程序从下面开始 assume cs:codesg,ds:data,s 09/15 11:12
- [genius023] 论坛里别人写的。。赞一个。差距啊 assume cs:codesg,ds:datasg,ss:st 09/15 10:26
[2008-10-07 21:23] 实验11
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[2008-09-21 21:05] 课程设计1
弄了一整天啊·····
结果是弄出来了,教训就是子程序里一定要多push,pop一点!
不过感觉有点乱。重新弄还得很久
不过。还是要庆祝一下~~~
assume cs:code
data segment
db '1975','1976','1977','1978','1979','1980','1981','1982', '1983'
db '1984', '1985','1986','1987','1988','1989','1990','1991','1992'
db '1993','1994','1995'
dd 16, 22, 382, 1356, 2...
结果是弄出来了,教训就是子程序里一定要多push,pop一点!
不过感觉有点乱。重新弄还得很久
不过。还是要庆祝一下~~~
assume cs:code
data segment
db '1975','1976','1977','1978','1979','1980','1981','1982', '1983'
db '1984', '1985','1986','1987','1988','1989','1990','1991','1992'
db '1993','1994','1995'
dd 16, 22, 382, 1356, 2...
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[2008-09-20 21:51] 实验10.3
现在会自己跟踪程序发现错误改过来了。
有进步,鼓励一下。嘿嘿
assume cs:code
data segment
db 10 dup(0)
data ends
code segment
start:
mov ax,12666
mov bx,data
mov ds,bx
mov si,0
call dtoc
mov dh,8
mov dl,3
mov cl,2
mov si,0;再复0
call show_str
mov ax,4c00h
int 21h
dtoc:
push di
push cx
push bx
mov di,0 ;用...
有进步,鼓励一下。嘿嘿
assume cs:code
data segment
db 10 dup(0)
data ends
code segment
start:
mov ax,12666
mov bx,data
mov ds,bx
mov si,0
call dtoc
mov dh,8
mov dl,3
mov cl,2
mov si,0;再复0
call show_str
mov ax,4c00h
int 21h
dtoc:
push di
push cx
push bx
mov di,0 ;用...
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[2008-09-20 13:20] 实验10.2
这个把公式意思搞明白了就做得挺顺
assume cs:code
code segment
start:
mov ax,4240h
mov dx,000fh
mov cx,0ah
call divdw
mov ax,4c00h
int 21h
divdw:
push bx
push ax ;低16位先存起来
mov ax,dx ;先把数据的高16位当被除数
mov dx,0;除数按16位算,所以被除数要32位。
div cx
mov bx,ax ;把得到的商存起来,余数还在dx中
pop ax;低16出马了
div cx
mov cx,dx;...
assume cs:code
code segment
start:
mov ax,4240h
mov dx,000fh
mov cx,0ah
call divdw
mov ax,4c00h
int 21h
divdw:
push bx
push ax ;低16位先存起来
mov ax,dx ;先把数据的高16位当被除数
mov dx,0;除数按16位算,所以被除数要32位。
div cx
mov bx,ax ;把得到的商存起来,余数还在dx中
pop ax;低16出马了
div cx
mov cx,dx;...
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[2008-09-20 08:29] 实验10.1
周末来啦
一大早起来真清静,,昨天晚上没弄通的,一下搞定了
assume cs:code
data segment
db 'welcome to masm!',0
data ends
code segment
start:
mov dh,8
mov dl,3
mov cl,2
mov ax,data
mov ds,ax
mov si,0
call show_str
mov ax,4c00h
int 21h
show_str:
mov ax,0b800h
mov es,ax
mov al,160
mul dh
mov bx,ax
sub ...
一大早起来真清静,,昨天晚上没弄通的,一下搞定了
assume cs:code
data segment
db 'welcome to masm!',0
data ends
code segment
start:
mov dh,8
mov dl,3
mov cl,2
mov ax,data
mov ds,ax
mov si,0
call show_str
mov ax,4c00h
int 21h
show_str:
mov ax,0b800h
mov es,ax
mov al,160
mul dh
mov bx,ax
sub ...
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[2008-09-15 09:16] 实验九
assume cs:code
data segment
db'w e l c o m e t o m a s m !'字母之间都有空格。。因为下面复制bx是加2。
data ends
code segment
start:
mov ax,data
mov ds,ax
mov ax,0b872h ;这前面要加个0!b8720h是屏幕中间的起始位置
mov es,ax
mov bx,0
mov si,0
mov cx,16
s: ;第一行
mov al,[bx]
mov es:[bx+si],al
mov al,0...
data segment
db'w e l c o m e t o m a s m !'字母之间都有空格。。因为下面复制bx是加2。
data ends
code segment
start:
mov ax,data
mov ds,ax
mov ax,0b872h ;这前面要加个0!b8720h是屏幕中间的起始位置
mov es,ax
mov bx,0
mov si,0
mov cx,16
s: ;第一行
mov al,[bx]
mov es:[bx+si],al
mov al,0...
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[2008-09-13 16:22] 实验七
assume cs:code
data segment
db '1975', '1976', '1977', '1978', '1979', '1980', '1981', '1982', '1983'
db '1984', '1985', '1986', '1987', '1988', '1989', '1990', '1991', '1992'
db '1993', '1994', '1995'
dd 16, 22, 382, 1356, 2390, 8000, 16000, 2446, 50065, 97479, 140417, 19...
data segment
db '1975', '1976', '1977', '1978', '1979', '1980', '1981', '1982', '1983'
db '1984', '1985', '1986', '1987', '1988', '1989', '1990', '1991', '1992'
db '1993', '1994', '1995'
dd 16, 22, 382, 1356, 2390, 8000, 16000, 2446, 50065, 97479, 140417, 19...
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[2008-09-12 22:07] 实验六
assume cs:code,ds:data,ss:stack
data segment
db '1. display '
db '2. brows '
db '3. replace '
db '4. modify '
data ends
stack segment
dw 0,0,0,0,0,0,0,0
stack ends
code segment
start: mov ax,data
mov ds,ax
mov bx,0
mov...
data segment
db '1. display '
db '2. brows '
db '3. replace '
db '4. modify '
data ends
stack segment
dw 0,0,0,0,0,0,0,0
stack ends
code segment
start: mov ax,data
mov ds,ax
mov bx,0
mov...
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[2008-09-11 20:59] 实验五(5)
assume cs:code
a segment
db 1,2,3,4,5,6,7,8
a ends
b segment
db 1,2,3,4,5,6,7,8
b ends
c segment
db 0,0,0,0,0,0,0,0
c ends
code segment
start:
mov ax,a
mov ds,ax
mov bx,0
mov cx,8
s:mov al,ds:[bx]
add al,ds:[bx+16]
mov ds:[bx+32],al
inc bx
loop s
mov ...
a segment
db 1,2,3,4,5,6,7,8
a ends
b segment
db 1,2,3,4,5,6,7,8
b ends
c segment
db 0,0,0,0,0,0,0,0
c ends
code segment
start:
mov ax,a
mov ds,ax
mov bx,0
mov cx,8
s:mov al,ds:[bx]
add al,ds:[bx+16]
mov ds:[bx+32],al
inc bx
loop s
mov ...
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[2008-09-09 20:51] 实验五 (6)
assume cs:code
a segment
dw 1,2,3,4,5,6,7,8,0ah,0ch
a ends
b segment
db 1,2,3,4,5,6,7,8
b ends
code segment
start:
mov ax,a
mov ds,ax
mov ax,b
mov ss,ax
mov sp,10h
mov bx,0
mov cx,8
s: mov ax,[bx]
...
a segment
dw 1,2,3,4,5,6,7,8,0ah,0ch
a ends
b segment
db 1,2,3,4,5,6,7,8
b ends
code segment
start:
mov ax,a
mov ds,ax
mov ax,b
mov ss,ax
mov sp,10h
mov bx,0
mov cx,8
s: mov ax,[bx]
...
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1