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Marcelo Tosatti writes:
>On Wed, 7 Mar 2001, Alexander Viro wrote:
>> You are reinventing the wheel.
>> man ptrace (see PTRACE_{PEEK,POKE}{TEXT,DATA} and PTRACE_{ATTACH,CONT,DETACH})
>
>With ptrace data will be copied twice. As far as I understood, Jeremy
>wants to avoid that.

Yes - I've been looking at the sys_ptrace code and it seems that it
does two copies through the kernel (but, using access_process_vm
instead of copy_from_user -- I'm not sure how they differ). I'm
trying to reduce the number of copies by giving one process a pointer
into another process's address space.

Right now, my code looks something like this: (it might make more
sense if you know that I've written a framework for writing user-space
device drivers... I'm going to be releasing it soon, hopefully after I
resolve this performance problem. Or maybe before, if it's hard.)

Process X: (any arbitrary process)
read(fd, my_local_buffer);
Kernel:
[notifies process Y that X needs data]
Process Y: (the userspace device driver)
memcpy(my_local_buffer, data_destined_for_process_x);
write(fd, my_local_buffer);
Kernel (in sys_write:)
copy_from_user(kernel_buffer, Y's my_local_buffer);
copy_to_user(X's my_local_buffer, kernel_buffer);

Now, this all works fine, but it's slower than I'd like. I used SGI's
kernprof (oss.sgi.com/projects/kernprof) and found that the kernel was
spending most of its time (like 70%) in copy_to_user and
copy_from_user. So, I am hoping to make this all go faster, while
hopefully preserving correctness, by changing to something like this:

Process X:
read(fd, my_local_buffer);
Kernel:
[maps X's my_local_buffer to Y's address space]
[notifies process Y that X needs data, and sends along a
pointer to write data to, or a handle usable by shmat,
or something]
Process Y:
memcpy(X's my_local_buffer, data_destined_for_process_x);

If I'm lucky we can avoid both copy_to_user and copy_from_user.

This would be easy if X was a "cooperative" process because we could
just set up a shared memory segment between the two processes. The
problem is that X is not cooperative -- it can be any existing
program.

Regards,
Jeremy
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