[157427] slow performance and eventual outofmemory error when revert with outline view open
diff --git a/bundles/org.eclipse.wst.xml.ui/src/org/eclipse/wst/xml/ui/internal/contentoutline/RefreshStructureJob.java b/bundles/org.eclipse.wst.xml.ui/src/org/eclipse/wst/xml/ui/internal/contentoutline/RefreshStructureJob.java
index 2fb90bb..1f6309e 100644
--- a/bundles/org.eclipse.wst.xml.ui/src/org/eclipse/wst/xml/ui/internal/contentoutline/RefreshStructureJob.java
+++ b/bundles/org.eclipse.wst.xml.ui/src/org/eclipse/wst/xml/ui/internal/contentoutline/RefreshStructureJob.java
@@ -1,5 +1,5 @@
/*******************************************************************************
- * Copyright (c) 2001, 2004 IBM Corporation and others.
+ * Copyright (c) 2001, 2006 IBM Corporation and others.
* All rights reserved. This program and the accompanying materials
* are made available under the terms of the Eclipse Public License v1.0
* which accompanies this distribution, and is available at
@@ -59,35 +59,49 @@
fRequests = new ArrayList(2);
}
- private synchronized void addRequest(Node node) {
+ private synchronized void addRequest(Node newNodeRequest) {
+ /*
+ * note: the caller must NOT pass in null node request (which, since
+ * private method, we do not need to gaurd against here, as long
+ * as we gaurd against it in calling method.
+ */
int size = fRequests.size();
- if (size > 0) {
- for (int i = 0; i < size; i++) {
- /*
- * If we already have a request which contains the new
- * request, discard the new request
- */
- Node node2 = (Node) fRequests.get(i);
- if (contains(node2, node))
- return;
- /*
- * If new request contains any existing requests, replace it
- * with new request
- */
- if (contains(node, node2)) {
- fRequests.set(i, node);
- return;
- }
- /**
- * None of the above, have a separate refresh subtree, then.
- */
- fRequests.add(node);
-
+ for (int i = 0; i < size; i++) {
+ Node existingNodeRequest = (Node) fRequests.get(i);
+ /*
+ * https://bugs.eclipse.org/bugs/show_bug.cgi?id=157427 If we
+ * already have a request which equals the new request, discard
+ * the new request
+ */
+ if (existingNodeRequest.equals(newNodeRequest))
+ return;
+ /*
+ * If we already have a request which contains the new request,
+ * discard the new request
+ */
+ if (contains(existingNodeRequest, newNodeRequest))
+ return;
+ /*
+ * If new request contains any existing requests, replace it with
+ * new request. ISSUE: technically, we should replace ALL
+ * contained, existing requests (such as if many siblings already
+ * que'd up when their common parent is then requested, but, I'm
+ * not sure if that occurs much, in practice, or if there's an
+ * algorithm to quickly find them all. Actually, I guess we could
+ * just go through the _rest_ of the list (i+1 to size) and remove
+ * any that are contained by new request ... in future :) .
+ */
+ if (contains(newNodeRequest, existingNodeRequest)) {
+ fRequests.set(i, newNodeRequest);
+ return;
}
}
- else {
- fRequests.add(node);
- }
+ /*
+ * If we get to here, either from existing request list being zero
+ * length, or no exisitng requests "matched" new request, then add the
+ * new request.
+ */
+ fRequests.add(newNodeRequest);
}
private synchronized void addViewer(StructuredViewer viewer) {
@@ -97,6 +111,9 @@
}
/**
+ * Simple hierarchical containment relationship. Note, this method returns
+ * "false" if the two nodes are equal!
+ *
* @param root
* @param possible
* @return if the root is parent of possible, return true, otherwise
