从 java 中的 list<String> 中删除一个值会抛出 java.lang.UnsupportedOperationException
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removing a value from a list<String> in java throws java.lang.UnsupportedOperationException
提问by Ashwin
I want to remove specific elements from my List. I don't want to do this while iterating through the list. I want to specify the value which has to be deleted. In javadocs I found the function List.remove(Object 0)This is my code :
我想从我的列表中删除特定元素。我不想在遍历列表时这样做。我想指定必须删除的值。在 javadocs 中，我找到了函数List.remove(Object 0)这是我的代码：
String str="1,2,3,4,5,6,7,8,9,10";
String[] stra=str.split(",");
List<String> a=Arrays.asList(stra);
a.remove("2");
a.remove("3");
But I get an Exception : java.lang.UnsupportedOperationException
但我得到一个例外： java.lang.UnsupportedOperationException
回答by NPE
The problem is that Arrays.asList()returns a list that doesn't support insertion/removal (it's simply a viewonto stra).
问题是Arrays.asList()返回一个不支持插入/删除的列表（它只是一个视图到stra）。
To fix, change:
要修复，请更改：
List<String> a = Arrays.asList(stra);
to:
到：
List<String> a = new ArrayList<String>(Arrays.asList(stra));
This makes a copy of the list, allowing you to modify it.
这会生成列表的副本，允许您对其进行修改。
回答by Eshan Sudharaka
http://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html#asList%28T...%29
http://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html#asList%28T...%29
See this. Arrays.asList returns a fixed list. Which is an immutable one. By its definition you cant modify that object once it creates.Thats why it is throwing unsupported exception.
看到这个。Arrays.asList 返回一个固定列表。这是一个不可变的。根据它的定义，一旦它创建，你就不能修改该对象。这就是它抛出不受支持的异常的原因。
