![[ASSEMBLY] Chia sẻ code một số bài toán](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEidI7Gd6Q8alj6ANcOhyma0f9FSWDt9NLMb1I4j-_Mzw_hxkWdvAvcS8pdsq_zc0UfPwR7glOlO50xdP_9N2lMYTu_fTfrAou91m-2BFqyxT5L88hV2kW7R8iT_NcF1pm41fbb0oBan874/s1600/school-assembly-color_t.gif "[ASSEMBLY] Chia sẻ code một số bài toán")
Máy tính đơn giản
Chương trình thực hiện các phép tính đơn giản như cộng trừ nhân chia. Phạm vi số: 16bit, số nguyên dương:
; ScanNum luu trong BX ScanNum MACRO LOCAL Nhap, KetThuc PUSH AX MOV BX, 0 Nhap: MOV AH, 1 INT 21h CMP AL, 13 JE KetThuc PUSH AX MOV AX, BX MUL ten MOV BX, AX POP AX AND AL, 0fh ADD BL, AL JMP Nhap KetThuc: POP AX ENDM ; Macro in text PrintText MACRO n PUSH AX PUSH DX MOV DL, n OR DL, 30h ; Chuyen so thanh ma ASCII tuong ung MOV AH, 2 INT 21h POP DX POP AX ENDM ; Chuogn tring chinh .MODEL SMALL ORG 100H .DATA MSG1 DB " My calculators", 0Ah, 0Dh, "$" MSG2 DB "Input a: $" MSG3 DB 0Ah, 0Dh, "Operator: (+, -, *, /): $" MSG4 DB 0Ah, 0Dh, "Input b: $" MSG5 DB 0Ah, 0Dh, "Result: $" num1 DW ? num2 DW ? oper DB ? ten DW 10 .CODE MOV AX, @DATA MOV DS, AX MOV AH, 9 LEA DX, MSG1 INT 21h LEA DX, MSG2 INT 21h ; Nhap so 1 ScanNum MOV num1, BX ; Print label MOV AH, 9 LEA DX, MSG3 INT 21h ; Nhap phep tinh MOV AH, 1 INT 21h MOV oper, AL ; Print label MOV AH, 9 LEA DX, MSG4 INT 21h ; Nhap so 2 ScanNum MOV num2, BX CMP oper, '+' JE PhepCong CMP oper, '-' JE PhepTru CMP oper, '*' JE PhepNhan JMP PhepChia PhepCong: MOV AX, num1 MOV BX, num2 ADD AX, BX JMP Exit PhepTru: MOV AX, num1 MOV BX, num2 SUB AX, BX JMP Exit PhepNhan: MOV DX, 0 MOV AX, num1 MOV BX, num2 MUL BX ; AX = AL * BL JMP Exit PhepChia: MOV DX, 0 MOV AX, num1 MOV BX, num2 DIV BX ; AX = AL / BL Exit: ; Print Label PUSH AX MOV AH, 9 LEA DX, MSG5 INT 21h POP AX ; Print res MOV CX, 1 ; Co danh dau MOV BX, 10000 Begin_Print: CMP BX, 0 JE End_Print ; BX = 0 CMP CX, 0 JE Calc ; CX = 0 CMP AX, BX JB Skip ; AX < BX Calc: MOV CX, 0 MOV DX, 0 DIV BX ; AX = DX AX / BX PrintText AL ; In AL nhu 1 ky tu binh thuong MOV AX, DX Skip: ; Giam BX 10 lan PUSH AX MOV DX, 0 MOV AX, BX DIV ten MOV BX, AX POP AX JMP Begin_Print End_Print: RET
Tìm ước chung lớn nhất của 2 số
Nhập vào 2 số 16bit rồi in ra UCLN của chúng:
; MACRO ; MACRO khong lam thay doi gia tri cua bien vao, chi thay doi gia tri thanh ghi Nhan MACRO reg, num LOCAL next, continue, exit PUSH AX PUSH DX PUSH CX MOV AX, reg MOV CX, num XOR DX, DX next: CMP CX, 0 JE exit TEST CX, 1b ; Ky tu cuoi cua AL = 1=> ZF = 0 JZ continue ; ZF = 1 ADD DX, AX continue: SHL AX, 1 SHR CX, 1 JMP next exit: MOV reg, DX POP CX POP DX POP AX ENDM ; Nhap so, luu vao AX ScanNum MACRO local Nhap, exit PUSH BX MOV AH, 1 XOR BX, BX ; Tuong duong mov bx, 0 Nhap: INT 21h CMP AL, 13 JE exit AND AL, 0Fh Nhan BX, 10 ADD BL, AL JMP Nhap exit: MOV AX, BX POP BX ENDM ; In 1 so bat ky o AX PrintNums MACRO LOCAL next, exit, print PUSH BX ; Cat cac gia tri vao stack PUSH CX PUSH DX MOV BX, 10 ; So chia XOR CX, CX ; MOV CX, 0 next: CMP AX, 0 JE exit ; Neu AX = 0 thi dung lai XOR DX, DX ; MOV DX, 0 DIV BX PUSH DX ; DX la so du cua phep chia INC CX ; Tang CX JMP next ; Lap lai exit: MOV AH, 2 print: XOR DL, DL ; MOV DL, 0 POP DX OR DL, 30h ; Chuyen so thanh ma ASCII tuong ung INT 21h LOOP print ; Lap den khi nao CX = 0 POP DX POP CX POP BX ENDM ; In so co 1 chu so PrintNum MACRO reg PUSH AX PUSH DX MOV DL, reg MOV AH, 2 OR DL, 30h INT 21h POP DX POP AX ENDM ; Tim uoc chung lon nhat, xuat ket qua vao DX UCLN MACRO m, n PUSH AX PUSH BX MOV AX, m MOV BX, n next: CMP AX, BX JE exit JB less SUB AX, BX JMP next less: SUB BX, AX JMP next exit: MOV DX, AX POP BX POP AX ENDM ; Chuong trinh chinh .MODEL small ORG 100h .DATA Enter DB 0Dh, 0Ah, '$' num1 DW ? num2 DW ? .CODE MOV AX, seg Enter MOV DS, AX ScanNum ; Xuat ra AL MOV num1, AX PUSH AX MOV AH, 9 LEA DX, Enter INT 21h POP AX ScanNum MOV num2, AX PUSH AX MOV AH, 9 LEA DX, Enter INT 21h POP AX XOR DX, DX UCLN num1, num2 MOV AX, DX PrintNums
Chương trình in ra n số hạng đầu tiên trong dãy Fibonacy
Nhập vào 1 số n (từ 1~22) rồi in ra n số hạng đầu tiên trong dãy Fibonacy
; MACRO ; MACRO khong lam thay doi gia tri cua bien vao, chi thay doi gia tri thanh ghi Nhan MACRO reg, num LOCAL next, continue, exit PUSH AX PUSH DX PUSH CX MOV AX, reg MOV CX, num XOR DX, DX next: CMP CX, 0 JE exit TEST CX, 1b ; Ky tu cuoi cua AL = 1=> ZF = 0 JZ continue ; ZF = 1 ADD DX, AX continue: SHL AX, 1 SHR CX, 1 JMP next exit: MOV reg, DX POP CX POP DX POP AX ENDM ; Nhap so, luu vao AX ScanNum MACRO local Nhap, exit PUSH BX MOV AH, 1 XOR BX, BX ; Tuong duong mov bx, 0 Nhap: INT 21h CMP AL, 13 JE exit AND AL, 0Fh Nhan BX, 10 ADD BL, AL JMP Nhap exit: MOV AX, BX POP BX ENDM ; In 1 so bat ky o AX PrintNums MACRO LOCAL next, exit, print PUSH BX ; Cat cac gia tri vao stack PUSH CX PUSH DX MOV BX, 10 ; So chia XOR CX, CX ; MOV CX, 0 next: CMP AX, 0 JE exit ; Neu AX = 0 thi dung lai XOR DX, DX ; MOV DX, 0 DIV BX PUSH DX ; DX la so du cua phep chia INC CX ; Tang CX JMP next ; Lap lai exit: MOV AH, 2 print: XOR DL, DL ; MOV DL, 0 POP DX OR DL, 30h ; Chuyen so thanh ma ASCII tuong ung INT 21h LOOP print ; Lap den khi nao CX = 0 POP DX POP CX POP BX ENDM .MODEL SMALL ORG 100h .DATA count DB ? Enter DB 0Ah, 0Dh, '?' ten DW 10 .CODE ScanNum MOV count, AL XOR DX, DX ; In dau Enter MOV AH, 2 MOV DL, 0Ah INT 21h MOV DL, 0Dh INT 21h CMP count, 1 JE in1 CMP count, 2 JE in11 OR BX, 1 PUSH BX PUSH BX ; In ra '1 1' MOV DL, '1' INT 21h MOV DL, ' ' INT 21h MOV DL, '1' INT 21h SUB count, 2 ; Dung stack de luu mang cac so Fibonacy next: MOV AH, 2 MOV DL, ' ' INT 21h MOV BP, SP MOV BX, [BP] ADD BX, [BP+2] PUSH BX MOV AX, BX PrintNums PUSH AX MOV AL, count DEC AL CMP AL, 0 JE exit MOV count, AL POP AX JMP next JMP exit ; In so Fibonacy in1: MOV DL, '1' INT 21h JMP exit ; In 2 so dau trong day in11: MOV DL, '1' INT 21h MOV DL, ' ' INT 21h MOV DL, '1' INT 21h exit:
