What will happen if 2 processes map same physical page

[Valdis Kl=?utf-8?Q?=c4=93?=tnieks]

On Wed, 20 Mar 2019 16:42:39 +0300, Lev Olshvang said:
> The question is it ipossiblle in Linux/MMU/TLB  that 2 processes map to
> the same physical address?

Totally possible.  That's how mmap shared memory works, and why shared
libraries are possible.

> Will CPU or  TLB discover that second process tries to reach occupied physical page?

Well, the hardware won't discover it as a "second" process, it only knows it's
processing *this* memory access.

> What if first process set page permission to read and second whats to write to this page ?

Perfectly OK - the two processes have separate page table mappings, with
separate permission bits. So (for example) physical page 0x17F000 is mapped to
virtual address 0x2034D000 with read-only permission n process 1's page tables,
and to virtual address 0x98FF3000 with read-write permission in process 2's
page tables. No problem.

(And before you ask, yes it's possible for process 2 to running on one core
doing a write to the page at the exact same time that process 1 is doing a read
on another core.  Depending on the hardware cache design, this may or may not
get process 1 updated data.  This is why locking and memory barriers are
important. See Documentation/memory-barriers.txt for more details)

"And then there's the Alpha" - a processor design that got much of its speed by
being weird about this stuff. :)

> Perhaps during context switch all page access permissions of first process is
> flashed out from MMU ?

Actually, the kernel just points the MMU at a new set of page table entries and lets
the TLB reload as needed. In particular, on most architectures, the kernel tries really
hard to ensure that all processes share at least part of their page table mappings so
the kernel is always mapped at the same place, meaning that there's a better chance
that on a syscall, the TLB already has hot entries for large parts of the kernel so no
TLB reloads are needed.