assume cs:code
code segment
mov ax,4240h
mov dx,000fh
mov cx,0ah
call divdw
mov ax,4c00h
int 21h
divdw: push ax
push dx
pop ax
mov dx,0
div cx
mov bx,ax
pop ax
div cx
mov cx,dx
mov dx,bx
...
- [zaixuexi] 我以为是a0=0,a1=1,没仔细看 02/12 09:23
- [yang5731] 不对吧 我的结果应该是对的 N=9时 确实得217 你自己算算。你帮我看看论坛 算法上的程序吧 题 02/11 23:49
- [zaixuexi] 递归不是这么写的哦,我想你肯定没调试过自己写的这代码. 先看C代码吧: unsigned int 02/11 22:58
- [游客] <a href="http://www.playren.com/">整人方法</a> 12/15 17:11
- [wisji8] mov ax,0a0h mul dh mov bx,ax mov ax,2 mul 10/03 16:56
- [yang5731] 谢谢了 课程设计一做出来了 确实不能调用着个····· 08/25 02:50
- [mywiil] 课程设计一要用到这个子程序,这里暂不做评论。 赶紧完成课程设计一吧。 08/21 03:33
[2010-08-19 08:20] 实验10.2
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[2010-08-19 08:20] 实验10.1
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: push dx
...
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: push dx
...
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[2010-08-13 07:27] 实验9
; 1行0a0H字节,第11行就是6e0H个字节,在0a0H个字节的中间显示
;20H个字节,就是从第40H个字节开始,那么就在720H处开始显示
assume cs:code
data segment
db 'welcome to masm!'
db 2h,24h,71h
data ends
stack segment
dw 8 dup (0)
stack ends
code segment
start:
mov ax,data
mov ds,ax
mov ax,stack
mo...
;20H个字节,就是从第40H个字节开始,那么就在720H处开始显示
assume cs:code
data segment
db 'welcome to masm!'
db 2h,24h,71h
data ends
stack segment
dw 8 dup (0)
stack ends
code segment
start:
mov ax,data
mov ds,ax
mov ax,stack
mo...
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[2010-08-07 02:08] 实验7 (用栈)
;有个疑惑,为什么用T命令执行pop [bx+2]后,
;Program terminated normally.程序好像直接运行完了,结果也正确
;------------------------------------------------------
assume cs:codesg
data segment
db '1975','1976','1977','1978','1979','1980','1981','1982','1983'
db '1984','1985','1986','1987','1988','1989','1990',...
;Program terminated normally.程序好像直接运行完了,结果也正确
;------------------------------------------------------
assume cs:codesg
data segment
db '1975','1976','1977','1978','1979','1980','1981','1982','1983'
db '1984','1985','1986','1987','1988','1989','1990',...
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[2010-08-06 22:27] 实验7
assume cs:codesg,ss:stacksg
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 16,22,382,1356,2390,8000,16000,24486,50065,97479,...
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 16,22,382,1356,2390,8000,16000,24486,50065,97479,...
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[2010-08-05 22:50] 实验6
assume cs:codesg,ss:stacksg,ds:datasg
stacksg segment
dw 0,0,0,0,0,0,0,0
stacksg ends
datasg segment
db '1. display '
db '2. brows '
db '3. replace '
db '4. modify '
datasg ends
codesg segment
start: mov ax,st...
stacksg segment
dw 0,0,0,0,0,0,0,0
stacksg ends
datasg segment
db '1. display '
db '2. brows '
db '3. replace '
db '4. modify '
datasg ends
codesg segment
start: mov ax,st...
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[2010-08-03 17:47] 实验5 求解
1.
(1) 0456,0123,0789,0abc,0def,0fde,0cba,0987
(2) 14d2 14d1 14d0
(3) X-2 X-1
2.
(1) 0456 ,0123
(2) 14d2 14d1 14d0
(3) X-2 X-1
3.
(1) 0456,0123
(2) 14d0 14d4 14d3
(3) X+3 X+4
4.
(3) 程序不不知道从哪里开始运行
5.
assume cs:code
a segment
db 1,2,3,4,5,6,7,8
a end...
(1) 0456,0123,0789,0abc,0def,0fde,0cba,0987
(2) 14d2 14d1 14d0
(3) X-2 X-1
2.
(1) 0456 ,0123
(2) 14d2 14d1 14d0
(3) X-2 X-1
3.
(1) 0456,0123
(2) 14d0 14d4 14d3
(3) X+3 X+4
4.
(3) 程序不不知道从哪里开始运行
5.
assume cs:code
a segment
db 1,2,3,4,5,6,7,8
a end...
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[2010-08-02 04:11] 实验4
(1)
assume cs:code
code segment
mov ax,0
mov ds,ax
mov bx,0
mov cx,64
s: mov [bx+200h],bx ;200后面一定要加H
inc bx
loop s
mov ax,4c00h
int 21h
code ends
end
(2)
code 18h
内存单元 CODE:0到0200:0
第2空先填CX
MOV...
assume cs:code
code segment
mov ax,0
mov ds,ax
mov bx,0
mov cx,64
s: mov [bx+200h],bx ;200后面一定要加H
inc bx
loop s
mov ax,4c00h
int 21h
code ends
end
(2)
code 18h
内存单元 CODE:0到0200:0
第2空先填CX
MOV...
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[2010-08-01 23:26] 实验3
(3) PSP中存放着文件名...
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[2010-08-01 21:51] 检测点2.3
4次
MOV AX,BX; IP+2
SUB AX,AX; IP+2 AX=0
JMP AX; 相当于MOV IP,0 这时IP+2 指令结束时IP=0...
MOV AX,BX; IP+2
SUB AX,AX; IP+2 AX=0
JMP AX; 相当于MOV IP,0 这时IP+2 指令结束时IP=0...
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