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style: Simplify :dir() implementation

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This I noticed while working on the following patches. Shouldn't have
any behavior change: the behavior does in fact match the element state
flag semantics correctly if we do this. We did split the dir flags into
two element bits a while ago.

:not(:dir()) still behaves correctly of course, and we have tests for that.

Differential Revision: https://phabricator.services.mozilla.com/D130734
This commit is contained in:
Emilio Cobos Álvarez 2023-05-31 11:37:02 +02:00 • committed by Oriol Brufau
parent 327812e3eb
commit 5b68241958
4 changed files with 6 additions and 34 deletions
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8
components/style/gecko/selector_parser.rs
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@ -162,7 +162,7 @@ impl NonTSPseudoClass {
([$(($css:expr, $name:ident, $state:tt, $flags:tt),)*]) => {
match *self {
$(NonTSPseudoClass::$name => flag!($state),)*
NonTSPseudoClass::Dir(..) |
NonTSPseudoClass::Dir(ref dir) => dir.element_state(),
NonTSPseudoClass::MozLocaleDir(..) |
NonTSPseudoClass::Lang(..) => ElementState::empty(),
}
@ -186,10 +186,8 @@ impl NonTSPseudoClass {
self.state_flag().is_empty() &&
!matches!(
*self,
// :dir() depends on state only, but doesn't use state_flag
// because its semantics don't quite match. Nevertheless, it
// doesn't need cache revalidation, because we already compare
// states for elements and candidates.
// :dir() depends on state only, but may have an empty
// state_flag for invalid arguments.
NonTSPseudoClass::Dir(_) |
// :-moz-is-html only depends on the state of the document and
// the namespace of the element; the former is invariant