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[TB Optimization] Skip subtrees based on the subtree's root node's permissions #4008

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4 changes: 4 additions & 0 deletions src/borrow_tracker/tree_borrows/perms.rs
Original file line number Diff line number Diff line change
Expand Up @@ -236,6 +236,10 @@ impl Permission {
pub fn is_active(&self) -> bool {
self.inner == Active
}
/// Check if `self` is the never-allow-writes-again state of a pointer (is `Frozen`).
pub fn is_frozen(&self) -> bool {
self.inner == Frozen
}

/// Default initial permission of the root of a new tree at inbounds positions.
/// Must *only* be used for the root, this is not in general an "initial" permission!
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26 changes: 25 additions & 1 deletion src/borrow_tracker/tree_borrows/tree.rs
Original file line number Diff line number Diff line change
Expand Up @@ -173,7 +173,7 @@ impl LocationState {
rel_pos: AccessRelatedness,
) -> ContinueTraversal {
if rel_pos.is_foreign() {
let new_access_noop = match (self.latest_foreign_access, access_kind) {
let mut new_access_noop = match (self.latest_foreign_access, access_kind) {
// Previously applied transition makes the new one a guaranteed
// noop in the two following cases:
// (1) justified by `foreign_read_is_noop_after_foreign_write`
Expand All @@ -185,6 +185,30 @@ impl LocationState {
// need to be applied to this subtree.
_ => false,
};
if self.permission.is_disabled() {
// A foreign access to a `Disabled` tag will have almost no observable effect.
// It's a theorem that `Disabled` node have no protected initialized children,
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That's not an obvious theorem -- can you give a brief argument?

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It's proven in Coq 😛.

The reason it holds is that to become disabled, you need to have a foreign write access happen. But that would have triggered any protected initialized nodes that are children of the node being disabled. And you can't have a new child of Disabled become initialized, because that would mean the to-be-initialized node has a child access, which is however blocked by the Disabled parent.

// and so this foreign access will never trigger any protector.
// Further, the children will never be able to read or write again, since they
// have a `Disabled` parents. Even further, all children of `Disabled` are one
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The argument could end here, right? The permissions below don't matter since anyway no access is possible.

// of `ReservedIM`, `Disabled`, or a not-yet-accessed "lazy" permission thing.
// The two former are already invariant under all foreign accesses, and for
// the latter it does not really matter, since they can not be used/initialized
// due to having a protected parent. So this only affects diagnostics, but the
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should be "disabled parent", right?

// blocking write will still be identified directly, just at a different tag.
new_access_noop = true;
}
if self.permission.is_frozen() && access_kind == AccessKind::Read {
// A foreign read to a `Frozen` tag will have almost no observable effect.
// It's a theorem that `Frozen` nodes have no active children, so all children
// already survive foreign reads. Foreign reads in general have almost no
// effect, the only further thing they could do is make protected `Reserved`
// nodes become conflicted, i.e. make them reject child writes for the further
// duration of their protector. But such a child write is already rejected
// because this node is frozen. So this only affects diagnostics, but the
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Is it possible to add a testcase that demonstrates the effect on diagnostics?

// blocking read will still be identified directly, just at a different tag.
new_access_noop = true;
}
if new_access_noop {
// Abort traversal if the new transition is indeed guaranteed
// to be noop.
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