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layout: Use the style of the nearest common ancestor node for background colors

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This commit is contained in:
Patrick Walton 2012-11-11 17:12:30 -08:00
parent 8c3b8fe3d4
commit 6d4cb4319d
1 changed files with 11 additions and 7 deletions
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18
src/servo/layout/box.rs
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@ -412,8 +412,8 @@ impl RenderBox : RenderBoxMethods {
match *self {
UnscannedTextBox(*) => fail ~"Shouldn't see unscanned boxes here.",
TextBox(_,d) => {
let nearest_element = self.nearest_element();
let color = nearest_element.style().color().to_gfx_color();
let nearest_ancestor_element = self.nearest_ancestor_element();
let color = nearest_ancestor_element.style().color().to_gfx_color();
list.append_item(~DisplayItem::new_Text(&abs_box_bounds,
~d.run.serialize(),
d.range,
@ -441,12 +441,16 @@ impl RenderBox : RenderBoxMethods {
self.add_border_to_list(list, &abs_box_bounds);
}
fn add_bgcolor_to_list(list: &mut DisplayList, abs_bounds: &Rect<Au>) {
fn add_bgcolor_to_list(@self, list: &mut DisplayList, abs_bounds: &Rect<Au>) {
use std::cmp::FuzzyEq;
if !self.d().node.is_element() { return }
// FIXME: This causes a lot of background colors to be displayed when they are clearly not
// needed. We could use display list optimization to clean this up, but it still seems
// inefficient. What we really want is something like "nearest ancestor element that
// doesn't have a RenderBox".
let nearest_ancestor_element = self.nearest_ancestor_element();
let bgcolor = self.style().background_color();
let bgcolor = nearest_ancestor_element.style().background_color();
if !bgcolor.alpha.fuzzy_eq(&0.0) {
list.append_item(~DisplayItem::new_SolidColor(abs_bounds, bgcolor.to_gfx_color()));
}
@ -535,8 +539,8 @@ impl RenderBox : BoxedDebugMethods {
// Other methods
impl RenderBox {
/// Returns the nearest ancestor-or-self node. Infallible.
fn nearest_element(@self) -> Node {
/// Returns the nearest ancestor-or-self element node. Infallible.
fn nearest_ancestor_element(@self) -> Node {
let mut node = self.d().node;
while !node.is_element() {
match NodeTree.get_parent(&node) {