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