--- title: "8086 ALP: max and min of 16-bit list" subject: "Microprocessor" chapter: 3 key: "semester_4/microprocessor:alp-8086-max-min-16bit" source: https://www.bctengineer.com/q/semester_4/microprocessor/alp-8086-max-min-16bit site: BCT Engineering --- # 8086 ALP: max and min of 16-bit list Microprocessor · Chapter 3: Programming with 8086 Microprocessor - Asked in: 2070 Magh, 8 marks - Video: [Assembly Language Programs of 8086- Multiplication, Division, Biggest, Smallest](https://www.youtube.com/watch?v=fcCswkm1R_Q) — Deepak Verma - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/microprocessor/alp-8086-max-min-16bit - Whole chapter: https://www.bctengineer.com/md/semester_4/microprocessor/chapter-3 ## Question Write a program for 8086 to find the largest and smallest value from a list of ten 16 bit data and store the result in maxval and minval variables. ## Short answer For ten 16-bit words, initialize AX max and BX min with first word, compare remaining nine unsigned, increment SI by 2, store in MAXVAL and MINVAL. Sample result MAXVAL=0A5A5H, MINVAL=0010H. ## Answer **Method:** load the first word into AX (maximum) and BX (minimum), then compare the other nine words (unsigned) with them, updating AX and BX. Finally store them in `MAXVAL` and `MINVAL`. Words occupy 2 bytes, so SI is incremented by 2. ```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 NUMS DW 1234H, 0056H, 8000H, 0FFFH, 4321H DW 0A5A5H, 0010H, 7FFFH, 2222H, 9000H MAXVAL DW ? MINVAL DW ? .CODE MAIN PROC MOV AX,@DATA MOV DS,AX LEA SI,NUMS ; SI -> list of ten words MOV AX,[SI] ; AX = maximum so far MOV BX,AX ; BX = minimum so far ADD SI,2 MOV CX,9 ; nine remaining words NEXT: MOV DX,[SI] ; DX = next word CMP DX,AX JBE CHKMIN ; DX <= max MOV AX,DX ; new maximum CHKMIN: CMP DX,BX JAE SKIP ; DX >= min MOV BX,DX ; new minimum SKIP: ADD SI,2 ; next word LOOP NEXT MOV MAXVAL,AX ; store results MOV MINVAL,BX MOV AH,4CH ; return to DOS INT 21H MAIN ENDP END MAIN ``` **Result:** MAXVAL = 0A5A5H and MINVAL = 0010H for the sample list. For signed 16-bit data, replace JBE with JLE and JAE with JGE. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does SI increase by 2 after each comparison? A. Each word occupies 2 bytes B. LOOP adds 2 automatically C. The array has 2 columns D. CMP uses two bytes of the next element Answer: A. The data are 16-bit words, so consecutive elements are 2 bytes apart. (Key point: Pointer step) 2. Which registers hold the running maximum and minimum? A. AX = minimum, BX = maximum B. AX = maximum, BX = minimum C. CX = maximum, DX = minimum D. Both are kept only in memory Answer: B. AX is loaded with the first word as maximum and BX with the same value as minimum. (Key point: Registers used) 3. What are MAXVAL and MINVAL for the sample list (unsigned)? A. MAXVAL = 9000H, MINVAL = 0010H B. MAXVAL = 7FFFH, MINVAL = 0010H C. MAXVAL = 0A5A5H, MINVAL = 0056H D. MAXVAL = 0A5A5H, MINVAL = 0010H Answer: D. Unsigned, 0A5A5H is the largest and 0010H the smallest of the ten words. (Key point: Result for sample) 4. How must the program change for signed 16-bit data? A. Use ADD SI,4 B. Replace JBE with JLE and JAE with JGE C. Replace CMP with TEST D. Use byte registers instead Answer: B. Signed comparison needs the signed jumps JLE/JGE; JBE/JAE treat 8000H as a large positive number. 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