CODEWITHSUNDEEP
linux-kerneloperating-system
user855
user855

Reputation: 19928

Writing to a disk without syscall

I am trying to understand how ring 3 to ring 0 transfer works in operating systems.

I think I understand how a syscall works.

My understanding is that when user mode program wants to make a syscall it will setup the call arguments and send an INT that will transfer over control to the OS which will then read the args, do that work and then return control back to user program. There Are more optimized sys enter arms sys exit variants as well.

All this makes sense to me If the user voluntarily calls the syscall.

However, to gurarantee safety OS cannot assume that callers will use syscall to access resources.

My question is — what happens if user program directly tries to access resource (disk) directly without involving OS.

How does the OS intercept it?

Upvotes: 2

Views: 920

Answers (1)

user15187356
user15187356

Reputation: 837

Any piece of I/O hardware, such as the disk controller, will (designer's choice) either respond to an I/O port address or a memory-space address, or possibly both. There is no other way to talk to the hardware. The hardware is sitting out on some bus. Program code must read/write some I/O port or must read/write some "memory" address which is really the device rather than actual RAM.

On x86, since the kernel controls access to both:

  • I/O ports, by setting or not setting the I/O port permissions, preventing ring 3 access

  • physical memory-space addresses (by controlling the virtual-to-physical address mapping)

then it can absolutely remove access from user mode.

So there is no instruction that user mode can execute that addresses the device. This is the fundamental aspect of the kernel/user split on any hardware: the kernel can control what user mode can do.

To pick up on a comment by @sawdust - once the kernel has set up the above restrictions, then:

  • an attempt to issue an I/O port instruction will trap to the kernel because access has not been granted.

  • access to memory-space device addresses is simply inexpressible; there is no user-space virtual address that equates to the particular physical address required.

Upvotes: 2

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