Bài tập Lập trình hợp ngữ - Số 3

Bài tập Lập trình hợp ngữ - Số 3 Bài tập: ; nhap 0<=x1, x2 , x3 <= 9 ; tinh tong, hieu, tich, thuong ; tim max, min; .model small .stack 100h .data tb1 db 10,'Nhap x1: $' tb2 db 10,'Nhap x2: $' tb3 db 10,'Nhap x3: $' x1 db ? x2 db ? x3 db ? tb4 db 10,'Tong: x1+x2+x3= $' tb5 db 10,'Hieu: x1-x2= $' tb6 db 10,'Tich: x1*x2= $' tb7 db 10,'Thuong: x1/x2= $' tong db 0 hieu db 0 .code ; lay dia chi doan du lieu mov ax, @data mov ds, ax ; xuat thong bao1 mov dx, offset tb1 call SubXuat ;Nhap x1 call SubNhap mov x1,al ; xuat thong bao2 mov dx, offset tb2 call SubXuat ;Nhap x2 call SubNhap mov x2,al ; xuat thong bao3 mov dx, offset tb3 call SubXuat ;Nhap x3 call SubNhap mov x3,al ; xuat thong bao4 mov dx, offset tb4 call SubXuat ; tinh tong call SubTong mov dx, offset tb5 call SubXuat ; tinh hieu call SubHieu ;call SubTich ;call SubThuong ;call subMax ;call SubMin ; tro ve dos mov ah, 4ch int 21h ;----------------------Xuat thong bao----------------- SubXuat PROC pop bx ; luu dia chi de quay ve mov ah, 9h int 21h push bx ret SubXuat ENDP ;------------------------Nhap Ky tu----------------- SubNhap PROC pop bx ; luu dia chi de quay ve mov ah, 1h int 21h push bx ret SubNhap ENDP ;-----------------------Tong x1+x2+x3---------------- SubTong PROC pop bx ; luu dia chi de quay ve mov al,x1 sub al,'0' mov tong, al mov al,x2 sub al,'0' add al, tong mov tong, al ; tong = x1+ x2 mov al,x3 sub al,'0' add al, tong ; tong = x1+x2+x2 cmp al, 14h jae HienHon20 cmp al, 0ah jae HienHon10 jmp HienTong HienHon20: ; luu lai al mov tong, al ; xuat ky tu 2 mov ah, 2h mov dl, '2' int 21h mov al, tong ; gan tong cho al sub al, 14h add al, 30h mov dl, al int 21h jmp ketthuc HienHon10: ; luu lai al mov tong, al ; xuat ky tu 2 mov ah, 2h mov dl, '1' int 21h mov al,tong ; gan tong cho al sub al, 0ah add al, 30h mov dl, al int 21h jmp ketthuc HienTong: add al, 30h mov ah, 2h mov dl, al int 21h ketthuc: push bx ret SubTong ENDP ;-----------------------Hieu x1-x2------------------- SubHieu Proc mov al, x2 sub al, '0' mov hieu, al mov al, x1 sub al, '0' sub al, hieu ; tinh x1-x2 cmp al, 0h ; so sanh voi 0 jl XuatAm add al, 30h mov dl, al mov ah, 2h int 21h jmp KThuc XuatAm: ; xuat dau - mov hieu, al ; luu al vao hieu mov dl, '-' mov ah, 2h int 21h ; xuat al mov al, 0ffh sub al, hieu add al, 31h mov dl, al mov ah, 2h int 21h KThuc: ret SubHieu Endp ;-----------------------Tich x1*x2------------------- ;SubTich Proc ;ret ;SubTich endp ;-----------------------Thuong x1/x2----------------- ;SubThuong Proc ;ret ;SubThuong endp ;-----------------------Max ------------------------- ;SubMax PROC ;ret ;SubMax ENDP ;---------------------------------------------------- ;SubMin PROC ;ret ;SubMin endP ;===================================================== end