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主题 : : 【实验5】第五小题两种解法 [待解决] |
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[帖 主] [ 发表时间:2010-01-09 12:01 ] | |
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信誉值:0
注册日期:2009-12-19 01:51 |
1, 利用三个段空间,DS,ES,SS来实现将a段和b段之和存入c段的目的
D:\ASM>type lab55.asm
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 ax,b
mov es,ax
mov ax,c
mov ss,ax
mov sp,0
mov bx,0
mov cx,8
s: mov ax,[bx]
add ax,es:[bx]
mov ss:[bx],ax
inc bx
loop s
mov ah,4ch
int 21
code ends
end start
D:\ASM>
实验结果:
-d ds:0 7
14D9:0000 01 02 03 04 05 06 07 08 ........
-d es:0 7
14DA:0000 01 02 03 04 05 06 07 08 ........
-d ss:0 7
14DB:0000 00 00 00 00 00 00 00 00 ........
程序结束后:
-d ds:0 7
14D9:0000 01 02 03 04 05 06 07 08 ........
-d es:0 7
14DA:0000 01 02 03 04 05 06 07 08 ........
-d ss:0 7
14DB:0000 02 04 06 08 0A 0C 0E 10 ........
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第二种方法, 利用两个短, 来实现目的。通过观察,a段和b段相连,所以呢, 可以把ab两段合二为一, 然后利用偏移,使程序精简了些
D:\ASM>
D:\ASM>
D:\ASM>
D:\ASM>type lab55a.asm
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 ax,c
mov es,ax
mov bx,0
mov cx,8
s: mov ax,[bx]
add ax,[bx+10h]
mov es:[bx],ax
inc bx
loop s
mov ah,4ch
int 21
code ends
end start
D:\ASM>
实验结果:
AX=14C6 BX=0000 CX=004F DX=0000 SP=0000 BP=0000 SI=0000 DI=0000
DS=14C4 ES=14C6 SS=14C4 CS=14C7 IP=000D NV UP EI PL NZ NA PO NC
14C7:000D B90800 MOV CX,0008
-d ds:0 1f
14C4:0000 01 02 03 04 05 06 07 08-00 00 00 00 00 00 00 00 ................
14C4:0010 01 02 03 04 05 06 07 08-00 00 00 00 00 00 00 00 ................
-d es:0 7
14C6:0000 00 00 00 00 00 00 00 00 ........
-g 1b
AX=0010 BX=0008 CX=0000 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000
DS=14C4 ES=14C6 SS=14C4 CS=14C7 IP=001B NV UP EI PL NZ NA PO NC
14C7:001B B44C MOV AH,4C
-t
AX=4C10 BX=0008 CX=0000 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000
DS=14C4 ES=14C6 SS=14C4 CS=14C7 IP=001D NV UP EI PL NZ NA PO NC
14C7:001D CD15 INT 15
-p
AX=8610 BX=0008 CX=0000 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000
DS=14C4 ES=14C6 SS=14C4 CS=14C7 IP=001F OV UP EI NG NZ NA PE CY
14C7:001F 7405 JZ 0026
-d ds:0 1f
14C4:0000 01 02 03 04 05 06 07 08-00 00 00 00 00 00 00 00 ................
14C4:0010 01 02 03 04 05 06 07 08-00 00 00 00 00 00 00 00 ................
-d es:0 7
14C6:0000 02 04 06 08 0A 0C 0E 10 ........
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[第19楼] [ 回复时间:2010-07-05 21:15 ] | |
荣誉值:0
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注册日期:2010-06-08 22:23 |
我写的程序,运行是没有问题,但是没有想到还有这么多简单的方法,学习了
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
cc segment
db 0, 0, 0, 0, 0, 0, 0, 0
cc ends
code segment
start: mov ax, a
mov ds, ax
mov bx, 0
mov cx, 8
s: sub dx, dx
mov ax, b
mov es, ax
mov dl, es:[bx]
mov al, ds:[bx]
add dl, al
mov ax, cc
mov es, ax
mov es:[bx], dl
inc bx
loop s
mov ax, 4c00h
int 21h
code ends
end start | | |
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