Arrays.asList引起的惨案

最近代码中需要对两个数组求交，想当然便用到了List中的retainAll函数，但要将将数组转换成list。代码如下：

String[] abc = new String[] { "abc", "acd", "add" };
String[] abd = new String[] {"acd", "cd", "de"};

List<String> abcList = Arrays.asList(abc);
List<String> abdList = Arrays.asList(abd);

abcList.retainAll(abdList);

没想到执行后却抛出了异常：

java.lang.UnsupportedOperationException
at java.util.AbstractList.remove(AbstractList.java:144)
at java.util.AbstractList$Itr.remove(AbstractList.java:360)
at java.util.AbstractCollection.retainAll(AbstractCollection.java:370)

反编译后才发现原来 Arrays.asList返回的ArrayList并不是java.util.ArrayList，而是在Arrays类中重新定义的一下内部类ArrayList：

private static class ArrayList<E> extends AbstractList<E>
implements RandomAccess, java.io.Serializable
{
private static final long serialVersionUID = -2764017481108945198L;
private final E[] a;

ArrayList(E[] array) {
if (array==null)
throw new NullPointerException();
a = array;

｝

//.....其它的省略

}

这个ArrayList并没有重写remove方法。

修改成如下，问题解决：

String[] abc = new String[] { "abc", "acd", "add" };
String[] abd = new String[] {"acd", "cd", "de"};

List<String> abcList = new ArrayList<String>(Arrays.asList(abc));
List<String> abdList = new ArrayList<String>(Arrays.asList(abd));

abcList.retainAll(abdList);

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Arrays.asList引起的惨案

原文：http://www.cnblogs.com/BensonHe/p/3604278.html