public int completed(R request) {
assert request.equals(requests.peek());
requests.poll();
int remaining = requests.size();
if (remaining != 0) coordinator.send(requests.peek());
return remaining;
}
public static void main(String[] args) {
HashSet hs = new HashSet();
hs.add(new R(5));
hs.add(new R(-3));
hs.add(new R(9));
hs.add(new R(-2));
// 打印HashSet集合，集合元素没有重复
System.out.println(hs);
// 取出第一个元素
Iterator it = hs.iterator();
R first = (R) it.next();
// 为第一个元素的count实例变量赋值
first.count = -3; // ①
// 再次输出HashSet集合，集合元素有重复元素
System.out.println(hs);
// 删除count为-3的R对象
hs.remove(new R(-3)); // ②
// 可以看到被删除了一个R元素
System.out.println(hs);
System.out.println("hs是否包含count为-3的R对象？" + hs.contains(new R(-3))); // 输出false
System.out.println("hs是否包含count为-2的R对象？" + hs.contains(new R(-2))); // 输出false
}
/**
* To find in the union, find in the components, concatenate the results, and omit duplicates.
* That last is a performance penalty, but I see no way to remove it unless we know the graphs do
* not overlap.
*/
@Override
public ExtendedIterator<Triple> graphBaseFind(final TripleMatch t) {
Set<Triple> seen = CollectionFactory.createHashedSet();
return recording(L.find(t), seen).andThen(rejecting(R.find(t), seen));
// return L.find( t ) .andThen( rejecting( R.find( t ), L ) );
}
@Override
public boolean graphBaseContains(Triple t) {
return L.contains(t) || R.contains(t);
}
/** To remove a triple, remove it from <i>both</i> operands. */
@Override
public void performDelete(Triple t) {
L.delete(t);
R.delete(t);
}
