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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.
Addressing mode of 8086 : Addressing mode specify a way of locating operands or data. Depending on the data types used the memory addressing modes and in the instruction ,
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
Program : A program to move a string of the data words from offset 2000H to offset 3000H the length of the string is OFH. Solution : For writing this program, we will use
I need some guidance on which project to make in assembly language
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Will be needing help with assembly language assignments over the course of 4 weeks
Description: LC3 allows input from keyboard and output to display on the screen. This lab will exercise the input/output capability using LC-3 Assembly language. Procedure
a- Trace the following program fragment and find out the content of ax after the the execution of the program. X db 5,7 -3,-9,4,-7,9 Mov
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