我有一个树状的结构,我的拷贝构造函数似乎偶尔会掉落一些“叶子”。
基本结构:
public class Arrow {
ArrayList<Arrow> subArrows;
Interval start;
Interval end;
}
复制构造函数:
public Arrow(Arrow other) {
this.start = new Interval(other.start);
this.end = new Interval(other.end);
if (other.subArrows != null) {
this.subArrows = new ArrayList<Arrow>();
for (Arrow sub : other.subArrows) {
this.subArrows.add(new Arrow(sub));
}
} else {
this.subArrows = new ArrayList<Arrow>();
}
}
我预计这基本上会对我的树结构进行深层复制。相反,我偶尔会发现我的一个subArrows
数组是空的。我没有注意到一种模式,除了它们往往是在我树的最低“水平”。
有什么想法吗?我有一段时间没用过java。
编辑:有几个人要求提供更多代码,所以这里有触及subArrows的所有地方。这是一个非常大的算法/数据结构,所以发布所有这些都是不合理的。递归获取所有subArrows并返回一组它们。
Set<Arrow> allSubArrows(Arrow arrow) {
Set<Arrow> arrowSet = new HashSet<Arrow>();
if (arrow.subArrows != null && arrow.subArrows.size() > 0) {
for (Arrow sub : arrow.subArrows) {
arrowSet.addAll(allSubArrows(sub));
}
return arrowSet;
} else {
arrowSet.add(arrow);
return arrowSet;
}
}
Mathy背后的原因,但在底部修改了subArrows:
void enforceMonotonicity(Arrow arrow) {
boolean changed = false;
if (arrow.end != null && arrow.start != null) {
if (arrow.start.isParallelTo(arrow.end)) {
//either left to right or bottom to top
if (arrow.start.isVertical()) {
//left interval pointing to right interval
if (arrow.start.startGraph.y > arrow.end.startGraph.y) {
arrow.end.startGraph.y = arrow.start.startGraph.y;
changed = true;
}
if (arrow.end.endGraph.y < arrow.start.endGraph.y) {
arrow.start.endGraph.y = arrow.end.endGraph.y;
changed = true;
}
} else {
//bottom interval pointing to top interval
if (arrow.start.startGraph.x > arrow.end.startGraph.x) {
arrow.end.startGraph.x = arrow.start.startGraph.x;
changed = true;
}
if (arrow.end.endGraph.x < arrow.start.endGraph.x) {
arrow.start.endGraph.x = arrow.end.endGraph.x;
changed = true;
}
}
}
}
if (changed) {
//check to make sure SOMETHING is still reachable, if not arrow = null
if (arrow.start.isVertical()) {
if (arrow.start.startGraph.y >= arrow.start.endGraph.y ||
arrow.end.startGraph.y >= arrow.end.endGraph.y) {
arrow = null;
}
} else {
if (arrow.start.startGraph.x >= arrow.start.endGraph.x ||
arrow.end.startGraph.x >= arrow.end.endGraph.x) {
arrow = null;
}
}
//if we changed the outer arrows, we need to recursively change the subarrows
if (arrow != null && arrow.subArrows != null && arrow.subArrows.size() > 0) {
for (Arrow sub : arrow.subArrows) {
enforceMonotonicity(sub);
}
}
}
}
合并算法的一部分:
HashSet<Arrow> mergeCells(Set<Arrow> first, Set<Arrow> second) {
HashSet<Arrow> mergedCell = new HashSet<Arrow>();
//loop through arrows in adjacent cells and find the ones that connect
for (Arrow topArrow : first) {
for (Arrow bottomArrow : second) {
if(topArrow.start.intersects(bottomArrow.end)) {
//merge arrows
Interval middle = topArrow.start.intersection(bottomArrow.end);
Arrow newArrow = new Arrow();
if (middle != null) {
//if they connect, we copy the two arrows, modify their connection,
//create a new arrow with the constituents as subarrows, and add that to the mergedcell
//after the mergedcell is created, we can delete the base arrows
Arrow topCopy = new Arrow(topArrow);
topCopy.start = middle;
Arrow bottomCopy = new Arrow(bottomArrow);
bottomCopy.end = middle;
newArrow.subArrows.add(topCopy);
newArrow.subArrows.add(bottomCopy);
newArrow.start = bottomCopy.start;
newArrow.end = topCopy.end;
//if end and middle are parallel, we need to project monotonicity
//monotonicity is already enforced within a single cell
//and enforcemonotonicity knows whether or not start and end are parallel
enforceMonotonicity(newArrow);
}
//enforceMonotonicity could null out the new arrow
if (newArrow != null && !newArrow.isNull()) {
mergedCell.add(newArrow);
}
}
}
}
//keep the originals in case they can connect later on
//(hashset doesn't allow duplicates, so no need to worry here)
mergedCell.addAll(first);
mergedCell.addAll(second);
return mergedCell;
}
答案 0 :(得分:0)
当(other.subArrows == null)
时,您的代码仍然会this.subArrows
创建new ArrayList<Arrow>()
。这不是一个真正的克隆(一个是null
,另一个是NOT)。如果你想确保subArrow总是非空(这有助于在不检查null
的情况下添加subArrow),你可以在字段定义中实例化它。
如果仍然无效,您可能需要显示更多代码,因为我们不知道如何在箭头中添加subArrows。此外,如果您的代码处于多线程环境下,并且该bug很难重现,则可能是同步问题。请注意,ArrayList
不是线程安全的。请尝试Vector
或Collections.synchronizedList
,或者正确同步关键块。同样,我们没有您的相关代码。
顺便说一句,声明List<Arrow> subArrows
优于ArrayList<Arrow> subArrows
。