style: Correct style sharing handling for any element that considered `:has()` in selector matching

For any element that anchors a `:has()` selector (i.e. Matches a selector that
contains a `:has()` on its rightmost side), we prevent style sharing altogether,
as evaluation of the `:has()` selector is required in the first place to
determine style sharing.

On the other hand, any element matching a rule containing `:has()` without
anchoring it can do style sharing for siblings, but not cousins.

Differential Revision: https://phabricator.services.mozilla.com/D176836

1 files changed, 9 insertions, 0 deletions

diff --git a/components/style/sharing/checks.rs b/components/style/sharing/checks.rs
index 42c7110dce8..a691ac5c764 100644
--- a/components/style/sharing/checks.rs
+++ b/components/style/sharing/checks.rs
@@ -36,6 +36,15 @@ where
 
     // Cousins are a bit more complicated.
     //
+    // The fact that the candidate is here means that its element does not anchor
+    // the relative selector. However, it may have considered relative selector(s)
+    // to compute its style, i.e. there's a rule `<..> :has(<..>) <..> candidate`.
+    // In this case, evaluating style sharing requires evaluating the relative
+    // selector for the target anyway.
+    if candidate.considered_relative_selector {
+        return false;
+    }
+
     // If a parent element was already styled and we traversed past it without
     // restyling it, that may be because our clever invalidation logic was able
     // to prove that the styles of that element would remain unchanged despite