--- title: "8086 ALP: largest and smallest, display" subject: "Microprocessor" chapter: 3 key: "semester_4/microprocessor:alp-8086-largest-smallest-display" source: https://www.bctengineer.com/q/semester_4/microprocessor/alp-8086-largest-smallest-display site: BCT Engineering --- # 8086 ALP: largest and smallest, display Microprocessor · Chapter 3: Programming with 8086 Microprocessor - Asked in: 2073 Bhadra, 8 marks - Video: [Execution of the 8086 Assembly Language Programming code To find the Largest and Smallest number.](https://www.youtube.com/watch?v=oP0tyaJQeyU) — Vaibhav KKM - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/microprocessor/alp-8086-largest-smallest-display - Whole chapter: https://www.bctengineer.com/md/semester_4/microprocessor/chapter-3 ## Question Write an assembly language program for 8086 to find the largest and smallest numbers of an array having 10 numbers. Display the found numbers in the clear screen. ## Short answer For array of 10 bytes, initialize largest in BL and smallest in BH with first element, compare remaining nine using unsigned JBE/JAE, update. Clear screen, display both in decimal. For sample array largest=120, smallest=3. ## Answer **Method:** the first element is taken as both the largest (BL) and the smallest (BH); the remaining nine are compared one by one and BL/BH are updated (unsigned comparison). The screen is cleared and both values are displayed in decimal with the helper `PRINT_DEC`. ```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 ARR DB 25,7,98,64,3,120,45,88,17,60 MSG1 DB 'Largest = $' MSG2 DB 'Smallest = $' CRLF DB 13,10,'$' .CODE MAIN PROC MOV AX,@DATA MOV DS,AX LEA SI,ARR ; SI -> array MOV BL,[SI] ; BL = largest (first element) MOV BH,BL ; BH = smallest INC SI MOV CX,9 ; 9 more elements NEXT: MOV AL,[SI] CMP AL,BL JBE NOTMAX ; AL <= largest MOV BL,AL ; new largest NOTMAX: CMP AL,BH JAE NOTMIN ; AL >= smallest MOV BH,AL ; new smallest NOTMIN: INC SI LOOP NEXT PUSH BX ; CLRSCR uses BH, so keep largest/smallest safe CLRSCR ; clear screen POP BX DISPSTR MSG1 MOV AL,BL XOR AH,AH CALL PRINT_DEC ; show largest DISPSTR CRLF DISPSTR MSG2 MOV AL,BH XOR AH,AH CALL PRINT_DEC ; show smallest MOV AH,4CH ; return to DOS INT 21H MAIN 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:** `Largest = 120` and `Smallest = 3`. (For signed numbers use JLE/JGE instead of JBE/JAE.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the largest and smallest values initialised? A. Both to 0 B. BL = 0 and BH = 0FFH C. Both to the first array element D. Largest to the last element, smallest to the first Answer: C. The first element is taken as both largest (BL) and smallest (BH), and the remaining nine are compared. (Key point: Initialisation) 2. Why is CX set to 9 for a 10-element array? A. The first element is already loaded, so nine remain to compare B. One element is reserved for the result C. The array index starts at 9 D. LOOP always runs one extra time Answer: A. After loading element 1 and incrementing SI, only 9 elements are left. (Key point: Loop count) 3. Which jumps are used for the unsigned comparison, and what changes for signed numbers? A. JBE/JAE; use JLE/JGE for signed B. JLE/JGE; use JBE/JAE for signed C. JE/JNE; use JZ/JNZ for signed D. JBE/JAE for both signed and unsigned Answer: A. JBE/JAE test the carry flag for unsigned data; JLE/JGE test sign and overflow for signed data. (Key point: Comparison type) 4. Why is BX pushed before CLRSCR? A. CLRSCR uses BH, which would destroy the smallest value B. CLRSCR uses BL only C. The stack must be cleared first D. INT 10h needs BX on the stack Answer: A. The CLRSCR macro loads BH with the attribute and later page number, overwriting the saved smallest value. 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