Int n-unconditional branch instruction-microprocessor, Assembly Language

INT N : Interrupt Type N:-

In the interrupt structure of 8086/8088, 256 interrupts are distinct equivalent to the types from OOH to FFH. When an instruction INT N is executed, the TYPE byte N is multiplied by value 4 and the contents of IP and CS register of the interrupt  service routine will be taken from hexadecimal multiplication (Nx4) as offset address and 0000 as the segment address. In other terms, the multiplication of type N by value 4 (offset) points to a memory block in the 0000 segment, which have the IP and CS register values of the interrupt service routine.

For the execution of this instruction, the IF ought to be enabled.

Example :

Therefore the instruction INT 20H will find out the address of the interrupt service routine as follows:

INT       20H

Type* 4 = 20 * 4 = 80H

Pointer to CS and IP of the ISR is 0000: 0080 H

Given figure shows the arrangement of CS and IP  register addresses of the ISR in the interrupt vector table.

922_RET.jpg

Posted Date: 10/12/2012 4:06:18 AM | Location : United States







Related Discussions:- Int n-unconditional branch instruction-microprocessor, Assignment Help, Ask Question on Int n-unconditional branch instruction-microprocessor, Get Answer, Expert's Help, Int n-unconditional branch instruction-microprocessor Discussions

Write discussion on Int n-unconditional branch instruction-microprocessor
Your posts are moderated
Related Questions
8279 Keyword /Display Controller : Figure shows the structure of 8279 and its interface to the bus. Addressing is according to the table given below. CS        RD

Modes of 8254 :   Mode 0 (Interrupt on Terminal Count)-GATE which value is 1 enables counting and GATE  which value is 0 disables counting, and GATE put not effect on

1. Start your program at address $8500. To do this you need to inform the assembler, through the EQU and ORG assembler directives, that you want your program to start at $8500. Thi

AND: Logical AND: This instruction bit by bit ANDs the source operand that might be an immediate, or a memory location or register to the destination operand that might be a memor

Assembly Language: Inside the 8085, instructions are really stored like binary numbers, not a very good manner to look at them and very difficult to decipher. An assembler is

I have two homework assignments due in 10 hours for the x86 processor assembly language

Write a program on the assembly language to do the following: 1- Allocate array with 32bit 100 element 2- Prompt the user to enter the maximum or the upper bound of the rando

DAS: Decimal Adjust after Subtraction:- This instruction converts the result of subtraction operation of 2 packed BCD numbers to a valid BCD number. The subtraction operation has

As an instance of the normal priority mode, imagine that initially AEOI is equal to 0 and all the ISR and IMR bits are clear. Also consider that, as shown in given figure, requests

Sum of series of 10 numbers and store result in memory location total