长安大学考博-3324-微机原理与应用-程序代码

1求平均值，找最大小值

buf db 10dup(?)

Sum dw?

Aver db?

Max db ?

Min db?

Start: mov cx, 10

lea si, BUF

xor ax, ax

mov bl,[si]

mov max,bl

mov min,bl LoopAdd:

add al, [si]

adc ah, 0

cmp max,[si]

ja l1

cmp min,[si]

jb l2

jmp goon

l1:mov,max,[si]

jmp goon

l2: mov min,[si]

jmp goon

goon: inc si

loop LoopAdd

mov Sum, ax

mov bl, 10

div bl

mov Aver, al

2正负分组计算，计算每组数db array 100dup(?)

db plarray 100 dup(?)

db miarray 100 dup(?)

plus db 0

minus db 0

zero db 0

lea si,array

lea di,plarray

lea bx,miarray

xor dx,dx

mov cx ,100

cld

check: lodsb

or al,al

js x1

jz x2

inc plus

mov [di],al

inc di

jmp next

x1: inc minus

mov [bx],al

inc bx

jmp next

x2: inc zero

next:loop check

mov ah 4ch

int 21h

1.学生成绩，统计分个分数段的

人数。

Data segment

Data1 db 80 dup(?)

Data2 db 5 dup(0)

Data ends

Code segment

Assume cs:code,ds:data

start: push ds

Mov ax,0

Push ax

Mov ax,data

Mov ds, ax

Mov cx,80

lea si,data1

lea di,data2

again: mov al,[si]

cmp al,90

jc next1

inc byte ptr[di]

jmp sto

next1:cmp al. 80

jc next2

inc byte ptr[di+1]

jmp sto

next2: cmp al,70

jc next3

inc byte ptr[di+2]

jmp sto

next3: cmp al,60

jc next4

inc byte ptr[di+3]

jmp sto

next4: inc byte ptr[di+4]

sto: inc si

loop again

mov ah,4ch

int 21h

code ends

end start

2.100的16位无符号数冒泡排序

Mem dw 100 dup(?)

Start:Lea di,mem

Mov bl,99

Next1:mov si,di

Mov cl,bl

Next2:mov ax,[si]

Add si,2

Cmp ax,[si]

Jnc next3

Mov dx,[si]

Mov [si-2],dx

Mov [si],ax

Next3: dec cl

Jnz next2

Dec bl

Jnz next1

3.ASCII码转二进制

Masc DB ‘2’’6’’A’’1’

Mbin db 2 dup(?)

Error db ‘error,$’

Start: mov cl,4

Mov ch,cl

Lea si,masc

Cld

Xor ax,ax

Xor dx,dx

Next1: lods masc

And al,7fh

Cmp al,’0’

Jl error

Cmp al,’9’

Jg next2

Sub al,30h

Jmp short next3

Next2:cmp al,’A’

Jl error

Cmp al,’F’

Jg error

Sub al,37h

Jmp short next3

Error: lea dx, error

Mov ah 09h

Int 21h

Next3:or dl,al

Ror dx,cl

Dec ch

Jnz next1

Mov word ptr mbin,dx 返回DOS

4.二进制转ascll码

Buff dw 4fb6h

Ascc db 5 dup(?)

Start: mov cx,5

Lea di,ascc

Xor dx,dx

Mov ax,buff

Mov bx,0ah

Agin: div bx

Add dl,30h

Mov [di],dl

Inc di

And ax,ax

Jz sto

Mov dl,0

Loop agin

Sto: 退出DOS

5.统计不超过255的字符串长度，二进制转换成16进制显示String db’this a string’,0dh Length dw ?

Cr db 13h

Message db ‘the string is too long ,0dh,0ah,’$’

Main proc far

Start: call strlen

Mov dx length

Cmp dx,100h

Jb next1

Lea dx,message

Mov ah,09h

Int 21h

Jmp next2

Next1:mov dh,dl

Mov cl,4

Shr dl,cl

Cmp dl,9

Jbe lp

Add dl,7

Lp: add dl,30h

Mov ah ,6

Int 21h

Mov dl,dh

And dl,0fh

Cmp dl,9

Jbe lp1

Add dl,7

Lp1: add dl,30h

Mov ah ,6

Int 21h

Mov dl,’h’

Mov ah,6

Int 21h

Next2:退出DOS

Main endp

Strlen proc

Lea di, string

Mov cx,offffh

Mov al,cr

Mov ah,’$’

Cld

Again:inc cx

Cmp cx,100h

Ja

Cmp [di],ah

Je done

Scasb

Jne again

Done:mov length,cx

ret

Strlen endp

6 键盘输入的字符串在内存中

查找该字符

Table db’ABCDEF…’

String1 db’please enter a

string:’0dh,0ah,’$’

String2 db’wrong!’0dh,0ah,’$’

String3 db ‘ok!’$

String4 db ‘no!’’$’

Butffer db 40,?40 dup(?)

Tab-len equ 26 堆栈

Start: lea dx, string1

Mov ah,09h

Int 21h

Lea dx,buffer

Mov ah,0ah

Int 21h

Mov si,dx

Inc si

Mov bl,[si]

Mov bh,0

Inc si

Lea di, table

Mov cx,tab-len

Cmp cx,bx

Jnc goon

Lea dx, string2

Jmp exit

Goon:cld

Mov al,[si]

Scan:Repnz scasb

Jz match

Error: lea dx, string4

Jmp exit

Match: inc cx

Cmp cx,bx

Jc error

Push cx

Push si

Push di

Mov cx,bx

Dec di

Repz cmpsb

Pop di

Pop si

Pop cx

Jz found

Jcxz error

Jmp scan

Found: dec di

Lea dx,string3

Exit: mov ah,09h

Int 21h

7.从一个字符串中删去一个字符String db ‘expereience’

Length dw $_string

Key db’x’堆栈

Main proc far

Start: lea bx,string

Lea cx, length

Push bx

Push cx

Mov al,key

Call delchar

Mov ah ,4ch

Int 21h

Main endp

Delchar proc

Push bp

Mov bp,sp

Push si

Push di

Push cx

Cld

Mov si,[bp+4]

Mov cx,[si]

Mov di,[bp+6]

Repnz scasb

Jne done

Mov si,[bp+4]

Dec word ptr[si]

Mov si, di

Dec di

Rep movsb

Done: pop cx

Pop di

Popsi

Pop bp

Ret 4

Delchar endp

8．二进制转BCD码

Mbin dw 4Fb6h

Mbcd db 5 dup(?)

Start: mov ax,mbin

Xor dx,dx

Lea bx,mbcd

Mov cx,0ah

div cx

mov [bx],dl

mov dl,0

div cx

inc bx

mov [bx],dl

mov dl,0

div cx

inc bx

mov[bx],dl

mov dl,0

div cx

inc bx

mov[bx],dl

inc bx

mov[bx],al

inc bx

退回dos

9无符号数求最大

Array dw ***

Max dw ?

Min dw ?

Start:

Lea si, array

Mov cx,N

Cld

Lodsw

Mov Max,ax

Mov min,ax

Dec cx

Next:lodsw

Cmp ax,max

Ja L1

Cmp ax,min

Jb l2

Jmp goon

L1:mov max,ax

Jmp goon

L2:mov min,ax

Jmp goon

Goon:loop next

10有符号数求绝对值最大

Array dw ***

Max dw ?

Start:

Lea si, array

Mov cx,N

Cld

Lodsw

Cmp ax,0

Jl abs

Jmp abs1

Abs: neg ax

Abs1:Mov Max,ax

Dec cx

Next:lodsw

Cmp ax,0

Jl abs2

Jmp abs3

Abs2: neg ax

Abs3:Cmp ax,max

Ja L1

Jmp Goon

L1:mov max,ax

Jmp Goon

Goon:loop next

11对50个学生的成绩从大到小

排序，并统计大于等于60分的

学生成绩，并在屏幕上显示出

来。

score 50 dup(?)

Count db 0

lea di, score

Mov bl,49

Next1:mov si,di

mov cl,bl

next2 :mov al,[si]

cmp al,60

add si,1

cmp al,[si]

jnc next3

mov ah,[si]

mov [si-1],ah

mov[si],ah next3: dec cl

jnz next2

dec bl

jnz next1 Mov cx,50

lea si, count

L1:mov al,[si]

Cmp al,60

Jae x1

Jmp x2

X1:inc byte ptr[si]

X2: inc si

Loop l1 Display:Mov dl,count Mov dh,dl

Mov cl,4

Shr dl,cl

Cmp dl,9

Jbe lp

Add dl,7

Lp: add dl,30h

Mov ah 06h

Int 21h

Mov dl,dh

And dl,0fh

Cmp dl.,9

Jbe lp1

Add dl,7

Lp1: add dl,30h

Mov ah,06

Int 21h

Mov dl,’h’

Mov ah,6

Int 21h

mov ah,4ch

int 21h