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
.model small
.data
a db 09h
b db 02h
c dw ?
.code
main proc
mov ax, @data
mov ds, ax
mov al, a
mov bl, b
add al, bl
mov ah, 0
mov c, ax
int 3 ; Breakpoint interrupt
mov ax, 4C00h ; Exit program
int 21h
main endp
end main
In 8086 assembly programming, interrupts play a crucial role in handling various operations, from debugging to system calls. Two commonly used interrupts are INT 3h (Breakpoint Interrupt) and INT 21h (DOS Interrupt). While both involve interrupt handling, their functionalities and use cases are completely different.
In this blog post, we will explore the differences between INT 3h and INT 21h, their respective use cases, and practical examples to understand their behavior.
TL;DR
INT 3h is used for debugging; it stops execution and hands control to the debugger.
INT 21h is used for system services like displaying messages, reading input, and terminating the program.
Key Difference:INT 3h is a 1-byte instruction (CC), while INT 21h requires function numbers in AH to specify system calls.
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.
; Data Segment
data segment
a dw 0FFFFh ; Example value that will cause a carry
b dw 0001h ; Example value
c dw ? ; To store result
carry_flag db 0 ; To store carry (0 or 1)
data ends
; Code Segment
code segment
assume cs:code, ds:data
start:
mov ax, data
mov ds, ax ; Initialize data segment
mov ax, a ; Load first number
mov bx, b ; Load second number
add ax, bx ; Perform addition
mov c, ax ; Store result in 'c'
jc carry_occurred ; Jump if carry flag is set
mov carry_flag, 0 ; No carry, store 0
jmp end_program
carry_occurred:
mov carry_flag, 1 ; Store carry flag as 1
end_program:
int 3 ; Halt program
code ends
end start
In the realm of 8086 assembly language, understanding the nuances of data declaration is crucial. Two fundamental directives, DW (Define Word) and DB (Define Byte), play pivotal roles in allocating memory and storing data.
DW (Define Word)
Memory Allocation: Reserves 2 bytes (16 bits) of memory for the specified data.
Data Storage: Stores a 16-bit integer value.
Usage: Ideal for larger numerical values, addresses, or any data that requires 16 bits of representation.
Example:
data segment
number DW 0x1234 ; Allocates 2 bytes and stores 1234h (4660 decimal)
data ends
In this blog post, we’ll delve into the world of assembly language programming using the 8086 microprocessors. We’ll explore a practical example: creating an assembly program to determine the largest number from a given set of values.
This C++ program demonstrates how to find the largest element from a set of five integers using 8086 assembly instructions embedded directly into the code. The inline assembly helps perform comparisons and updates the result with the highest value found.
This post demonstrates how to sort an array of integers using inline assembly in C++. Here, we perform sorting in ascending order by comparing and swapping adjacent elements using embedded assembly within a C++ program.
#include<iostream.h>
#include<conio.h>
void main()
{
int a[5], x, y;
int i, j;
cout << "\n Enter 5 Numbers:";
for(i = 0; i < 5; i++)
{
cin >> a[i];
}
//Sorting
for(i = 0; i < 4; i++)
{
for(j = 0; j < 4; j++)
{
x = a[j];
y = a[j + 1];
_asm {
mov ax, x
mov bx, y
cmp ax, bx
jl nxt
mov cx, ax
mov ax, bx
mov bx, cx
mov x, ax
mov y, bx
}
nxt:
a[j] = x;
a[j + 1] = y;
}
}
cout << "\n Sorted Array:";
for(i = 0; i < 5; i++)
cout << a[i] << " ";
getch();
}
