What happens in an interrupt service routine?

Question

Can someone please explain to me what happens inside an interrupt service routine (although it depends upon specific routine, a general explanation is enough)? This always used

Answer 1

Minimal 16-bit example

The best way to understand is to make some minimal examples yourself.

First learn how to create a minimal bootloader OS and run it on QEMU and real hardware as I've explained here: https://stackoverflow.com/a/32483545/895245

Now you can run in 16-bit real mode:

    movw $handler0, 0x00
    mov %cs, 0x02
    movw $handler1, 0x04
    mov %cs, 0x06
    int $0
    int $1
    hlt
handler0:
    /* Do 0. */
    iret
handler1:
    /* Do 1. */
    iret

This would do in order:

• Do 0.
• Do 1.
• hlt: stop executing

Note how the processor looks for the first handler at address 0, and the second one at 4: that is a table of handlers called the IVT, and each entry has 4 bytes.

Minimal example that does some IO to make handlers visible.

Protected mode

Modern operating systems run in the so called protected mode.

The handling has more options in this mode, so it is more complex, but the spirit is the same.

Minimal example

See also

Related question: What does "int 0x80" mean in assembly code?