Question

Explain interrupts of 8086. Draw interrupt vector table

Answer 1

Interrupt vector table on 8086 is a vector that consists of 256 total interrupts placed at first 1 kb of memory from 0000h to 03ffh, where each vector consists of segment and offset as a lookup or jump table to memory address of bios interrupt service routine (f000h to ffffh) or dos interrupt service routine address, the call to interrupt service routine is similar to far procedure call.

The size for each interrupt vector is 4 bytes (2 word in 16 bit), where 2 bytes (1 word) for segment and 2 bytes for offset of interrupt service routine address.

So it takes 1024 bytes (1 kb) memory for interrupt vector table. On 8086 with dos operating system, interrupt vector table at 00h-1fh (int num 0-31) consists of lookup / jump table

address to hardware or bios interrupt handler routine, meanwhile 20h-ffh (int num 32-255) consist of jump table address to dos interrupt handler routine.

For example int 13h that located on ivt at 0000:004c contains address of Bios ROM Interrupt Service Routine, what it records is segment F000h, and offset 1140h, each bytes of that address will be placed little endian. The lower the interrupt number on interrupt vector table means the more priority needed for an interrupt.

Interrupt Vector Table (taken from somewhere on internet)

ivt