--- title: "8086 ALP: multiplication table of a number" subject: "Microprocessor" chapter: 3 key: "semester_4/microprocessor:alp-8086-multiplication-table" source: https://www.bctengineer.com/q/semester_4/microprocessor/alp-8086-multiplication-table site: BCT Engineering --- # 8086 ALP: multiplication table of a number Microprocessor · Chapter 3: Programming with 8086 Microprocessor - Asked in: 2068 Magh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/microprocessor/alp-8086-multiplication-table - Whole chapter: https://www.bctengineer.com/md/semester_4/microprocessor/chapter-3 ## Question Write a program in 8086 to read a number and display the multiplication table of that number. ## Short answer The program reads a decimal number using READ_NUM (accumulates digits until Enter), then loops i=1 to 10, printing "N x i = N*i" using MUL and PRINT_DEC. Output example: for 7, it displays 7 x 1 = 7 through 7 x 10 = 70. ## Answer **Method:** `READ_NUM` reads a multi-digit decimal number (digits are accumulated as value = value × 10 + digit until Enter is pressed). The program then prints `N x i = N*i` for i = 1 to 10. `MUL CX` gives N × i in DX:AX, and `PRINT_DEC` prints any value in AX in decimal. ```asm .MODEL SMALL .STACK 100H DISPSTR MACRO MSG ; display a '$'-terminated string LEA DX,MSG MOV AH,09H INT 21H ENDM CLRSCR MACRO ; clear screen and home the cursor MOV AX,0600H ; scroll whole window (AL=0 -> blank it) MOV BH,07H ; normal attribute MOV CX,0000H ; upper-left corner (row 0, col 0) MOV DX,184FH ; lower-right corner (row 24, col 79) INT 10H MOV AH,02H ; set cursor position MOV BH,00H ; page 0 MOV DX,0000H ; row 0, col 0 INT 10H ENDM .DATA MSG1 DB 'Enter a number: $' TMS DB ' x $' EQS DB ' = $' CRLF DB 13,10,'$' NUM DW ? .CODE MAIN PROC MOV AX,@DATA MOV DS,AX DISPSTR MSG1 CALL READ_NUM ; AX = number MOV NUM,AX DISPSTR CRLF MOV CX,1 ; multiplier i = 1 TBL: MOV AX,NUM CALL PRINT_DEC ; N DISPSTR TMS ; " x " MOV AX,CX CALL PRINT_DEC ; i DISPSTR EQS ; " = " MOV AX,NUM MUL CX ; DX:AX = N * i CALL PRINT_DEC ; product DISPSTR CRLF INC CX CMP CX,10 JBE TBL ; repeat for i = 1..10 MOV AH,4CH ; return to DOS INT 21H MAIN ENDP ; ---- read a decimal number from the keyboard, result in AX READ_NUM PROC PUSH BX PUSH CX PUSH DX XOR BX,BX ; BX = value accumulated so far RN1:MOV AH,01H ; read a character with echo INT 21H CMP AL,0DH ; Enter ends the number JE RN2 SUB AL,'0' ; ASCII -> digit XOR AH,AH MOV CX,AX ; CX = digit MOV AX,10 MUL BX ; AX = value * 10 ADD AX,CX ; add the new digit MOV BX,AX JMP RN1 RN2:MOV AX,BX ; return value in AX POP DX POP CX POP BX RET READ_NUM ENDP ; ---- print the unsigned number in AX in decimal (registers preserved) PRINT_DEC PROC PUSH AX PUSH BX PUSH CX PUSH DX MOV BX,10 XOR CX,CX ; digit counter PD1:XOR DX,DX DIV BX ; AX = quotient, DX = remainder (digit) PUSH DX INC CX OR AX,AX JNZ PD1 PD2:POP DX ; digits come out in correct order ADD DL,30H ; to ASCII MOV AH,02H INT 21H LOOP PD2 POP DX POP CX POP BX POP AX RET PRINT_DEC ENDP END MAIN ``` **Output for 7:** `7 x 1 = 7` ... `7 x 10 = 70`. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does READ_NUM build a multi-digit number? A. value = value + digit B. value = value x 10 + digit until Enter C. value = digit x 10 D. value = value / 10 Answer: B. Each new digit shifts the earlier digits left decimally, which is done by multiplying by 10 and adding. (Key point: Reading number) 2. What does MUL CX give when AX holds N? A. N + CX in AX B. N x CX in DX:AX C. N / CX in AX D. N x CX in BX Answer: B. A 16-bit MUL multiplies AX by the operand and places the 32-bit result in DX:AX. (Key point: Multiplication) 3. How is the table limited to 10 lines? A. CX starts at 1 and the loop repeats while CX <= 10 (CMP CX,10 / JBE) B. CX starts at 0 and loops until zero C. The user enters the limit D. The loop stops when AX overflows Answer: A. CX is the multiplier i, incremented each time, and JBE continues while i is 10 or below. (Key point: Loop range) 4. What is subtracted from the typed key code to get a digit? A. 'A' B. 30 C. 0DH D. '0' Answer: D. Subtracting ASCII '0' (30H) turns the key code into its value 0 to 9. 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