我不明白爲什麼這段代碼不會拋出一個NullPointerException後行我iterator.next();的JavaDoc說:預期NullPointerException異常與ConcurrentHashMap不顯示
視圖的iterator [..]確保遍歷元件,因爲它們 迭代器構造時的存在,並且可以(但不是 保證)反映構造後的任何修改。
在所有的運行中,我沒有反映任何修改,否則cur.getKey()會給我一個NullPointerException,因爲我刪除了一個元素。我甚至睡了一覺,以促進錯誤。
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map.Entry;
import java.util.concurrent.ConcurrentHashMap;
class MapWorker implements Runnable {
private final ConcurrentHashMap<Integer, String> map;
public MapWorker(final int i, final ConcurrentHashMap<Integer, String> map) {
this.map = map;
}
@Override
public void run() {
// Set key and value to search for
final Integer key = 3;
final String value = "three";
final Iterator<Entry<Integer, String>> iterator = map.entrySet().iterator();
System.out.println("Map is " + map);
while (iterator.hasNext()) {
try {
Thread.sleep(2000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
final Entry<Integer, String> cur = iterator.next();
if (cur.getKey()==key && cur.getValue().equals(value)) {
iterator.remove();
System.out.println("Removed");
}
}
}
}
public class MapExercise {
static final int NUM_WORKERS = 5;
public static void main(final String[] args) {
// Create and populate map
final ConcurrentHashMap<Integer, String> map = new ConcurrentHashMap<Integer, String>();
map.put(1, "one");
map.put(2, "two");
map.put(3, "three");
map.put(4, "four");
map.put(5, "five");
// Create and init worker threads
final List<Thread> allThreads = new ArrayList<Thread>();
for (int i = 0; i < NUM_WORKERS; i++)
allThreads.add(new Thread(new MapWorker(i, map)));
// Start worker threads
System.out.println("---------------------------");
for (final Thread t : allThreads)
t.start();
// Wait for worker threads to finish
for (final Thread t : allThreads)
try {
t.join();
} catch (final InterruptedException e) {
/* do nothing */
}
System.out.println("---------------------------");
System.out.println(map);
}
}
都說iterator.hasNext()返回true;同時(上下文切換)另一個線程刪除數字3,這應該是迭代器的下一個元素。上下文返回到已經進入while(iterator.hasNext())的前一個線程。 iterator.next()返回什麼?無論它返回什麼,我都會使用它,應該拋出異常的東西。 – Christian