c - Does int 80h interrupt a kernel process? -

first background knowledge, book: linux system programming: talking directly kernel , c library

signals mechanism one-way asynchronous notifications. signal may sent kernel process, process process, or process itself.

the linux kernel implements 30 signals.

signals interrupt executing process, causing stop whatever doing , perform predetermined action.

ok moving further, here quote part:

on intel family of microprocessors, such pentium, int 80h assembly language op code interrupt 80h. syscall interrupt on typical intel-based unix system, such freebsd. allows application programmers obtain system services unix kernel.

i can not quite make connection in head really. when example use

write  

method defined in posix, , when compiled assembly, , further assembled object code , linked executable in given architecture runs linux.... system call made right?

i assuming compiled code this:

mov eax,4 ;code system_write mov ebx,1 ;standard output mov ecx, memlocation; memlocation location number of ascii present mov edx, 20; 20 bytes written  int 80h; 

ok question @ point. int 80h send signal kernel / interrupt kernel? kernel one process? (is init process?) when cpu executes int 80h , happens? registers full of information already, (eax, ebx, ecx , edx in example..), how information used?

i can not quite make connection between cpu - kernel , cpu executes int 80h.

i can imagine code resides somewhere in memory sends required information device driver process code belong to? (i assuming kernel kernel 1 process?) , how int 80h instruction jump code? linux has implement somehow?

is kernel 1 process? (is init process?)

the kernel magic beast. it's not process. kernel doesn't have pid can refer to.

first, it's worth stating (even though it's obvious) instructions runs on processor: therefore, int 80h executed processor.

there called interrupt request handler. somehow similar function pointer. processor has table of interrupt handler. table called interrupt descriptor table (aka idt) , system wide (ie, not every process has it's own table). believe table populated kernel when first boot.

so, happens when int 80 executed?

1. the processor running in ring 3 protection level (the normal level process). more info on ring level, see this.
2. the processor switch ring 0, aka kernel mode. in mode, hardware protection disabled. mean code executed on can whatever wants. write everywhere in physical memory, rewrite interrupt descriptor table, etc.
3. the processor jump code located in interrupt descriptor table 80h interrupt. space available each interruption in idt small. why code jump again somewhere else.
4. the previous jump lend processor in kernel routine dedicated handling int 80h. processor no longer running process' code, running kernel code.
5. the kernel can check registers , memory , determine why interrupt triggered. understand wanted execute system call write.
6. the kernel code jump again, time in routine handle write. kernel runs code write.
7. the kernel done running code. tells processor go ring 3 protection level, , resume process.
8. userspace process (aka process) resumes.