Java ConcurrenLinkedHashMap迭代器提供的元素顺序取决于键大小?
我发现(迭代器?)ConcurrentLinkedHashMap的奇怪行为。 如果键长,则通过迭代entrySet/keySet获得的元素的顺序是奇怪的/意外的。如果钥匙短,一切都好 以下代码:Java ConcurrenLinkedHashMap迭代器提供的元素顺序取决于键大小?,java,guava,Java,Guava,我发现(迭代器?)ConcurrentLinkedHashMap的奇怪行为。 如果键长,则通过迭代entrySet/keySet获得的元素的顺序是奇怪的/意外的。如果钥匙短,一切都好 以下代码: public static void main(String[] args) { ConcurrentLinkedHashMap<String, String> map = new ConcurrentLinkedHashMap.Builder<
public static void main(String[] args) {
ConcurrentLinkedHashMap<String, String> map =
new ConcurrentLinkedHashMap.Builder<String, String>().maximumWeightedCapacity(1000).build();
for (int i = 0; i < 5; i++) {
map.put(i + "", i + "");
}
print(map);
map.clear();
// NOW AGAIN THE SAME, BUT WITH LONG KEY
for (int i = 0; i < 5; i++) {
map.put(i + "aaaaaaaaaaaaaaaaaaaaaaaaaa" +
"aaaaaaaaaaaaaaaaaaaaaaaaaa" +
"aaaaaaaaaaaaaaaaaaaaaaaaaa", i + "");
}
print(map);
}
private static void print(ConcurrentLinkedHashMap<String, String> a) {
Iterator iterator = a.entrySet().iterator();
while (iterator.hasNext()) {
System.out.println(" = " + iterator.next());
}
}
这很奇怪。
如果我使键变长,结果就不同了
是虫子吗?我怎样才能得到正确的结果?
(第二个“打印”结果的顺序应与第一个相同)
与java.util.LinkedHashMap
不同,此类不提供可预测的迭代顺序。
()
文档对此相当清楚-因此没有bug
关于迭代项的顺序,ConcurrentLinkedHashMap
提供了键的保留顺序。这可能并不总是您的目标,但至少对于给定的测试场景,在给定以下代码的情况下,返回一个有序的输出:
private static void print(ConcurrentLinkedHashMap<String, String> a)
{
Iterator iterator = a.entrySet().iterator();
while (iterator.hasNext())
{
System.out.println(" = " + iterator.next());
}
Iterator<String> iter = a.ascendingKeySet().iterator();
while (iter.hasNext())
{
String key = iter.next();
System.out.println(key + " -> " + a.get(key));
}
}
你用的是什么版本的番石榴?@assylias根据git的说法,它从来都不是番石榴的一部分,它只是为
MapMaker
和CacheBuilder
设计的。好吧,是的,谢谢。Strage为什么它被称为Linked*。@user\x更新了我的帖子,加入了一些关于ascendingKeySet
@user\x的信息。该类提供了LinkedHashMap的access order=true
功能,这在多线程中更为常见,并且难以扩展。相比之下,插入顺序并发映射相对来说微不足道。
private static void print(ConcurrentLinkedHashMap<String, String> a)
{
Iterator iterator = a.entrySet().iterator();
while (iterator.hasNext())
{
System.out.println(" = " + iterator.next());
}
Iterator<String> iter = a.ascendingKeySet().iterator();
while (iter.hasNext())
{
String key = iter.next();
System.out.println(key + " -> " + a.get(key));
}
}
= 0=0
= 1=1
= 2=2
= 3=3
= 4=4
0 -> 0
1 -> 1
2 -> 2
3 -> 3
4 -> 4
= 1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa=1
= 4aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa=4
= 2aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa=2
= 3aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa=3
= 0aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa=0
0aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa -> 0
1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa -> 1
2aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa -> 2
3aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa -> 3
4aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa -> 4