【正文】 lea dx,betterstr mov ah,09h int 21h add good,30h lea dx,goodstr mov ah,09h int 21h add pass,30h lea dx,passstr mov ah,09h int 21h add notpass,30h lea dx,notpassstr mov ah,09h int 21h 。給定一個學生成績，給出其名次 lea dx,Inputinfo mov ah,09h int 21h lea dx,inputscore mov ah,0ah int 21h mov cl,inputscore+1 cmp cl,2 ja S100 jb S1 mov al,inputscore+2 sub al,30h mov ch,10 mul ch add al,inputscore+3 sub al,30h jmp GoS100: mov al,100 jmp GoS1: cmp cl,0 jz S0 mov al,inputscore+2 sub al,30h jmp GoS0: mov al,0Go: mov bx,0 mov cx,5loop5: cmp al,score[bx] jae goon1 inc rankgoon1: inc bxloop loop5 add rank,30h lea dx,Rankinfo mov ah,09h int 21h mov ah,4ch int 21h cseg ends end Start 四． 實驗結果（包括必要的截圖）實驗三 匯編語言子程序設計(分支、子程序)一. 實驗目的掌握分支和子程序結構和設計方法；熟悉在PC機上建立、匯編、連接、調試和運行8086/8088匯編語言程序的過程。2. 轉換使用子程序實現，且：二進制到十進制和十六進制轉換使用內存?zhèn)鬟f參數；十進制到二進制和十六進制轉換使用寄存器傳遞參數；十六進制到二進制和十進制轉換使用堆棧傳遞參數。,0dh,0ah,39。,0dh,0ah,39。,0dh,0ah,39。,0dh,0ah,39。Msg1 db 0dh,0ah,39。Msg3 db 0dh,0ah,39。Msg5 db 0dh,0ah,39。Endpro db 0dh,0ah,39。 jz Do1 jmp Do2Do1: or bx,axDo2: shr ax,cl inc si cmp si,17 jbe Do mov ax,bx mov dx,0 mov cx,10000 div cx mov bx,dx and ax,00ffh add al,30h mov dl,al mov ah,2 int 21h mov ax,bx mov dx,0 mov cx,1000 div cx mov bx,dx and ax,00ffh add al,30h mov dl,al mov ah,2 int 21h mov ax,bx mov cl,100 div cl mov bh,ah add al,30h mov dl,al mov ah,2 int 21h mov al,bh mov ah,0 mov cl,10 div cl mov bh,ah add al,30h mov dl,al mov ah,2 int 21h add bh,30h mov dl,bh mov ah,2 int 21h pop dx pop cx pop bx pop ax pop si retBtoD endpBtoH proc push si push ax push bx push cx push dx lea dx,Msg1 mov ah,9 int 21h lea dx,buf mov ah,0ah int 21h lea dx,Msg6 mov ah,9 int 21h mov ax,8000h mov bx,0 mov cl,1 mov si,2Don: cmp buf[si],39。二. 實驗內容1 編寫一個宏，求三個數的最小數，原型為：MIN3,X,Y,Z,Min macro x,y,z local Continue,Change,Change2,Down push ax push dx mov al,x cmp al,y jg ChangeContinue: cmp al,z jg Change2 jmp DownChange: mov al,y jmp ContinueChange2: mov al,zDown: mov dl,al mov ah, 2 int 21h pop dx pop axendmdseg segment msg db 0dh,0ah,39。 p1 db ? p2 db ? p3 db ?dseg endscseg segment assume cs:cseg,ds:dsegStart: mov ax ,dseg mov ds,ax mov dx,offset msg mov ah,9 int 21h mov ah,1 int 21h mov p1,al mov ah,9 int 21h mov ah,1 int 21h mov p2,al mov ah,9 int 21h mov ah,1 int 21h mov p3,al mov dx,offset msg22 定時器 21的0BH功能intno equ 1ch data segment para public 39。 assume cs:code,ds:dataStart: mov ax,data mov ds,ax mov ax,0 mov es,ax cli mov ax,es:[intno*4] mov old1c[0],ax mov ax,es:[intno*4+2] mov old1c[2],ax sti cli mov word ptr es:[intno*4],offset TimerISR mov word ptr es:[intno*4+2],seg TimerISR stiWaithere: cmp isdone,1 jnz WaithereExit: cli mov ax,old1c[0]