从同一个数组创建两个列表，修改一个列表，更改另一个列表

Question

I created two list from same array and sorted one of them. When I tried to change one list, other list also got updated.

List<Integer> list = Arrays.asList(ar);
List<Integer> sorted = Arrays.asList(ar);
Collections.sort(sorted);
list.set(0,10000000); //changes sorted also

It took me a while figure out, below mentioned code worked.

List<Integer> sorted = new ArrayList<Integer>(Arrays.asList(ar));

I want to know why my first approach didn't work? I created two separate lists, why the changes are taking place in both of them. How does java assign values to variables here?

Answer 1

From the Java documentation for Arrays.asList :

Returns a fixed-size list backed by the specified array. (Changes to the returned list "write through" to the array.) This method acts as bridge between array-based and collection-based APIs, in combination with Collection.toArray() . The returned list is serializable and implements RandomAccess .

So when you change something in list, it "writes through" to the underlying array, ar , which is also the underlying array in sorted, so the change is reflected in sorted as well.

Also, the code for asList is:

public static <T> List<T> asList(T... a) {
    return new ArrayList<T>(a);
}

This is java.util.Arrays.ArrayList , which has the following definition:

ArrayList(E[] array) {
    a = Objects.requireNonNull(array);
}

What is important is that a is not copied, it is the original array. The java.util.ArrayList class has the following constructor

public ArrayList(Collection<? extends E> c) {
     elementData = c.toArray();
     size = elementData.length;
     // c.toArray might (incorrectly) not return Object[] (see 6260652)
     if (elementData.getClass() != Object[].class)
         elementData = Arrays.copyOf(elementData, size, Object[].class);
 }

so in the java.util.ArrayList constructor, we create copies of each element, and in java.util.Arrays.ArrayList , we do not.

Answer 2

Arrays有自己的ArrayList实现，它不会从toList复制数组

Answer 3

A List is a collection of objects, and both list are collections of the same objects. The set statement changes an object, and the object is shared by both lists.

I don't understand why the second version works.

Answer 4

list和sorted仍然指向第一种方法中的ar的相同内存地址，而在第二种方法中，在调用构造函数之后已经分配了新的内存地址，新的内存块被分配给类对象。