实验3 转移指令跳转原理及其简单应用编程
实验任务 1
- 给出程序task1.asm源码,及,运行截图
assume cs:code, ds:data
data segment
x db 1, 9, 3
len1 equ $ - x
y dw 1, 9, 3
len2 equ $ - y
data ends
code segment
start:
mov ax, data
mov ds, ax
mov si, offset x
mov cx, len1
mov ah, 2
s1:mov dl, [si]
or dl, 30h
int 21h
mov dl, ' '
int 21h
inc si
loop s1
mov ah, 2
mov dl, 0ah
int 21h
mov si, offset y
mov cx, len2/2
mov ah, 2
s2:mov dx, [si]
or dl, 30h
int 21h
mov dl, ' '
int 21h
add si, 2
loop s2
mov ah, 4ch
int 21h
code ends
end start
回答问题①
① line27, 汇编指令 loop s1 跳转时,是根据位移量跳转的。通过debug反汇编,查看其机
器码,分析其跳转的位移量是多少?(位移量数值以十进制数值回答)从CPU的角度,说明
是如何计算得到跳转后标号s1其后指令的偏移地址的。
位移量为:0x001B-0x000D=-14
IP=(IP)-14=0x001B-14
回答问题②
② line44,汇编指令 loop s2 跳转时,是根据位移量跳转的。通过debug反汇编,查看其机
器码,分析其跳转的位移量是多少?(位移量数值以十进制数值回答)从CPU的角度,说明
是如何计算得到跳转后标号s2其后指令的偏移地址的。
位移量为:0x0039-0x0029=-16
IP=(IP)-16=0x0039-16
③ 附上上述分析时,在debug中进行调试观察的反汇编截图
shown above
实验任务 2
给出程序 task2.asm 源码
assume cs:code, ds:data
data segment
dw 200h, 0h, 230h, 0h
data ends
stack segment
db 16 dup(0)
stack ends
code segment
start:
mov ax, data
mov ds, ax
mov word ptr ds:[0], offset s1
mov word ptr ds:[2], offset s2
mov ds:[4], cs
mov ax, stack
mov ss, ax
mov sp, 16
call word ptr ds:[0]
s1: pop ax
call dword ptr ds:[2]
s2: pop bx
pop cx
mov ah, 4ch
int 21h
code ends
end start
① 根据call指令的跳转原理,先从理论上分析,程序执行到退出(line31)之前,寄存器(ax) = s1的偏移地址
,寄存器(bx) = s2 的偏移地址
,寄存器(cx) = cs
。
② 对源程序进行汇编、链接,得到可执行程序task2.exe。使用debug调试,观察、验证调试结果与理论分析结果是否一致。
给出分析、调试、验证后,寄存器(ax) = 0021 (bx) = 0026 (cx) = 076C。与分析一致。
实验任务 3
给出程序源码task3.asm
assume cs:code, ds:data
data segment
x db 99, 72, 85, 63, 89, 97, 55
len equ $- x
data ends
stack segment
db 16 dup(0)
stack ends
code segment
start:
mov ax, stack
mov ss, ax
mov sp, 16
mov ax, 78
mov cx, len
mov ax, data
mov ds, ax
mov bx, 0
s1:
mov ah, 0
mov al, [bx]
call printNumber
call printSpace
inc bx
loop s1
mov ah, 4ch
int 21h
printNumber PROC NEAR
push bx
mov bh, 0ah
div bh
mov bx, ax
mov ah, 2
mov dl, bl
or dl, 30h
int 21h
mov ah, 2
mov dl, bh
or dl, 30h
int 21h
pop bx
ret
printNumber ENDP
printSpace PROC NEAR
mov ah, 2
mov dl, ' '
int 21h
ret
printSpace ENDP
code ends
end start
运行测试截图
实验任务 4
给出程序源码task4.asm
assume cs:code, ds:data
data segment
str db 'try'
len equ $ - str
data ends
stack segment
db 16 dup(0)
stack ends
code segment
start:
mov ax, stack
mov ss, ax
mov sp, 16
mov ax, 78
mov ax, data
mov ds, ax
mov si, 0
mov cx, len
mov bl, 2
mov bh, 0
call printStr
mov si, 0
mov cx, len
mov bl, 4
mov bh, 24
call printStr
mov ah, 4ch
int 21h
printStr PROC NEAR
mov ax, 0b800H
mov es, ax
mov di, 0
push cx
mov ch, 0
mov cl, bh
ps1:
add di, 160
loop ps1
pop cx
ps2:
mov al, [si]
mov es:[di], al
mov es:[di+1], bl
inc si
add di, 2
loop ps2
ret
printStr ENDP
code ends
end start
运行测试截图
实验任务 5
给出程序源码task5.asm
assume cs:code, ds:data
data segment
stu_no db '201983290307'
len = $ - stu_no
line db 80 dup(' ')
llen = $ - line
lineend = $
data ends
stack segment
db 16 dup(0)
stack ends
code segment
start:
mov ax, stack
mov ss, ax
mov sp, 16
mov ax, 78
mov ax, data
mov ds, ax
mov cx, 24
mov bh, 0
s1:
push cx
mov si, len
mov cx, llen
mov bl, 00010111b
call printStr
inc bh
pop cx
loop s1
mov ax, 80
sub ax, len
mov bh, 2
div bh
mov cl, al
mov ch, 0
mov si, len
s2:
mov byte ptr [si], '-'
inc si
loop s2
mov bx, 0
mov cx, len
s3:
mov al, [bx]
mov [si], al
inc bx
inc si
loop s3
s4:
mov byte ptr [si], '-'
inc si
cmp si, lineend
jnz s4
mov si, len
mov cx, llen
mov bl, 00010111b
mov bh, 24
call printStr
mov ah, 4ch
int 21h
printStr PROC NEAR
mov ax, 0b800H
mov es, ax
mov di, 0
push cx
mov ch, 0
mov cl, bh
ps1:
add di, 160
loop ps1
pop cx
ps2:
mov al, [si]
mov es:[di], al
mov es:[di+1], bl
inc si
add di, 2
loop ps2
ret
printStr ENDP
code ends
end start
运行测试截图