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SubjectRe: [GIT PULL] PM updates for 2.6.33
On Mon, 7 Dec 2009, Linus Torvalds wrote:

> > The consequence is that there's no way to hand off an entire subtree to
> > an async thread. And as a result, your single-pass algorithm runs into
> > the kind of "stall" problem I described before.
> No, look again. There's no stall in the thing, because all it really
> depends on is (for the suspend path) is that it sees all children before
> the parent (because the child will do a "down_read()" on the parent node
> and that should not stall), and for the resume path it depends on seeing
> the parent node before any children (because the parent node does that
> "down_write()" on its own node).
> Everything else is _entirely_ asynchronous, including all the other locks
> it takes. So there are no stalls (except, of course, if we then hit limits
> on numbers of outstanding async work and refuse to create too many
> outstanding async things, but that's a separate issue, and intentional, of
> course).

It only seems that way because you didn't take into account devices
that suspend synchronously but whose children suspend asynchronously.

A synchronous suspend routine for a device with async child suspends
would have to look just like your usb_node_suspend():

/* Wait until the children are suspended */
Suspend dev

/* Allow the parent to suspend */

So now suppose we've got two USB host controllers, A and B. They are
PCI devices, so they suspend synchronously. Each has a root hub child
(P and Q respectively) which is a USB device and therefore suspends
asynchronously. dpm_list contains: A, P, B, Q. (In fact A doesn't
enter into this discussion; you can ignore it.)

In your one-pass algorithm, we start with usb_node_suspend(Q). It does
down_read(B->lock) and starts an async task for Q. Then we move on to
suspend_one_node(B). It does down_write(B->lock) and blocks until the
async task finishes; then it suspends B. Finally we move on to
usb_node_suspend(P), which does down_read(A->lock) and starts an async
task for P.

The upshot is that P is stuck waiting for Q to suspend, even though it
should have been able to suspend in parallel. This is simply because P
precedes B in the list, and B is synchronous and must wait for Q to

With my two-pass algorithm, we start with Q. The first loop does
down_read(B->lock) and starts an async task for Q. We move on to B and
do down_read(B->parent->lock), nothing more. Then we move to to P,
with down_read(A->lock) and start an async task for P. Finally we do
down_read(A->parent->lock). Notice that now there are two async tasks,
for P and Q, running in parallel.

The second pass waits for Q to finish before suspending B
synchronously, and waits for P to finish before suspending A
synchronously. This is unavoidable. The point is that it allows P and
Q to suspend at the same time, not one after the other as in the
one-pass scheme.

Alan Stern

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