60 8086 microprocessor viva questions and answers for Sunday-evening exam prep, covering architecture, addressing modes, arithmetic and flags, the stack, interrupts and DOS/BIOS calls, each answer linked to a worked program or reference post.
This blog post details an 8086-assembly program that computes the power of a number using the Exponentiation by Squaring algorithm (O(log n) efficiency). While a standard iterative approach multiplies the base n times (taking O(n) time), exponentiation by squaring—also known as binary exponentiation—works by breaking the exponent down into its binary components. By squaring the base in each step and only multiplying it into the result when a bit in the exponent is set, we drastically reduce the computational load. For example, calculating x32 requires only 5 multiplications instead of 31. This example demonstrates advanced assembly concepts like bitwise manipulation, conditional branching, and efficient arithmetic optimization. Let’s get started! Logic Breakdown: The algorithm follows the mathematical identity of Binary Exponentiation: Check Exponent: If it's zero, stop. Odd Case: If the current exponent is odd, multiply the running result by the current base. Square and Halve: Regardless of odd/even, square the base and divide the exponent by 2. Loop: Continue until the exponent is exhausted. Let's visualize it's working for 53. StepBase ExponentResult ActionInitial531Start loopIter 153 (Odd)51 X 5 = 5Square251552 = 25, 3/2 = 1Iter 2251 (Odd)1255 X 25 = 125Square6250125252 = 625,1/2 = 0Exit-0125Loop terminates
This blog post details an 8086-assembly program that computes the power of a number, specifically baseexponent. This example demonstrates fundamental assembly concepts like loops, multiplication, and register manipulation to perform iterative arithmetic. Let’s get started! data segment base dw 0003h exponent dw 0004h result dd ? data ends code segment assume cs:code, ds:data start: ; Initialize data segment mov ax, data mov ds, ax ; Initialize result to 1 (R = B^0) mov word ptr result, 0001h mov word ptr result+2, 0000h ; Load base and exponent mov cx, exponent ; CX = exponent (loop counter) mov bx, base ; BX = base cmp cx, 0000h je exit ; If exponent is 0, result is already 1. Jump to exit. power_loop: ; Multiply result by base ; The result is a 32-bit number in result[0] (lower word) and result[2] (upper word) ; Multiplication requires careful handling of the 32-bit result with a 16-bit multiplier. ; 1. Multiply the lower word of result by base mov ax, word ptr result ; AX = result[0] mul bx ; AX * BX -> DX:AX (32-bit product) push dx ; Save the higher word (DX) of the product mov word ptr result, ax ; Store the lower word (AX) of the product as the new result[0] ; 2. Multiply the upper word of result by base mov ax, word ptr result+2 ; AX = result[2] mul bx ; AX * BX -> DX:AX (32-bit product) ; 3. Add the two 16-bit 'carry' terms pop cx ; Retrieve the saved higher word (DX) from step 1 into CX add ax, cx ; AX = AX + CX (Sum of two high words) adc dx, 0000h ; DX = DX + 0 + Carry from the previous ADD (final carry from the 32-bit multiplication) ; 4. Store the final upper word mov word ptr result+2, ax ; Store AX as the new result[2] ; Handle the 32-bit overflow (DX is the final carry from the 32-bit multiplication) ; For a 32-bit result storage, this program assumes the result fits in 32-bits. ; If the power is large, overflow might occur, which is a limitation of this 32-bit storage approach. loop power_loop ; Decrement CX and jump back to power_loop if CX != 0 exit: int 3 ; Program termination code ends end start
This blog post will walk you through a simple 8086 assembly program designed to perform bitwise operations: AND, OR, XOR and NOT on 8-bit numbers. While these operations have a straightforward purpose, they are foundational when working at low level, manipulating bits, flags and registers. Let’s get started! data segment a db 0Ah ; first 8-bit operand (10 decimal) b db 05h ; second 8-bit operand (5 decimal) res_and db ? ; result of a AND b res_or db ? ; result of a OR b res_xor db ? ; result of a XOR b res_not db ? ; result of NOT a data ends code segment assume cs:code, ds:data start: mov ax, data mov ds, ax ; load operands into registers mov al, a mov bl, b ; AND operation: AL = AL AND BL and al, bl mov res_and, al ; reload operands for next operation mov al, a mov bl, b ; OR operation: AL = AL OR BL or al, bl mov res_or, al ; reload operands for next operation mov al, a mov bl, b ; XOR operation: AL = AL XOR BL xor al, bl mov res_xor, al ; NOT operation on first operand: AL = NOT AL mov al, a not al mov res_not, al int 3 ; breakpoint / stop for debug code ends end start
In this blog post, we’ll explore an 8086 assembly program that computes the Greatest Common Divisor (GCD) of two numbers using the classic Euclidean Algorithm. The Euclidean method is an efficient and elegant way to find the largest number that divides both inputs without leaving a remainder. The principle is simple:GCD(A, B) = GCD(B, A mod B) This process repeats until the remainder becomes zero. At that point, the divisor is the GCD. Let’s dive into the code and understand how it works! Program Code data segment a dw 0012h ; First number (18 decimal) b dw 000Ah ; Second number (10 decimal) gcd_result dw ? ; Variable to store the GCD data ends code segment assume cs:code, ds:data start: mov ax, data mov ds, ax ; Initialize data segment mov ax, a ; Load A into AX mov bx, b ; Load B into BX cmp bx, 0000h ; Check if B is zero jz store_result gcd_loop: mov dx, 0000h ; Clear DX before division div bx ; AX = AX / BX, remainder in DX cmp dx, 0000h ; If remainder = 0, GCD found je store_result_bx mov ax, bx ; A = B mov bx, dx ; B = remainder jmp gcd_loop ; Repeat loop store_result_bx: mov ax, bx ; Move GCD into AX store_result: mov gcd_result, ax ; Store final result int 3 ; Stop execution code ends end start
In this post, we'll walk through an 8086 assembly program that calculates the factorial of a given integer. This program highlights the power of the 8086's loop instruction and the use of general-purpose registers for iterative calculations. We'll use the AX register as an accumulator for the result and the CX register as a loop counter, which is its special-purpose function. Let's get to the code!
The complete starting point for 8086 study: the processor's 1978 origins, why Intel built it, its headline features, real-world applications, and the memory segmentation mechanism that lets 16-bit registers address a full 1 MB — with the physical address formula, all four segment registers, overlap rules, .COM vs .EXE differences, and the .MODEL directive.
This blog post will guide you through a MASM (Microsoft Macro Assembler) program that performs the addition of two 8-bit numbers. While this is a fundamental operation, it demonstrates crucial assembly programming concepts such as data handling, register usage, and memory storage. Let’s dive in! Assembly Code
INT 3 is the one-byte CC debugger trap; INT 21h is the two-byte CD 21 gateway to DOS, where AH selects the service. Covers the opcode difference (NASM's int 3 is not CC), the stack frame, DEBUG behaviour, the standard AH function table from Ralf Brown's list, a runnable companion module and 14 viva questions.
In this blog post, we’ll explore an 8086 assembly language program that adds two 16-bit numbers while also checking for a carry flag.