日期:2014-05-20  浏览次数:20881 次

求教一个把list转换成数组的问题!!!!
public class ListToArrayTest {
public static void main(String args[]) {
ArrayList<String> list = new ArrayList<String>();
String group[];
list.add("aaa");
list.add("bbb");
list.add("ccc");
list.add("ddd");

// 1
// Object[] temp;
// temp = list.toArray();
// System.out.println(temp.length);
// group=new String[temp.length];
// for(int i=0;i<temp.length;i++){
// group[i]=(String) temp[i];
// }
//  
// 2
  group =new String[list.size()];
  group= (String[]) list.toArray();

  //Object to String
   
   
  System.out.println("group size is "+group.length);
}
}


报错的是:
Exception in thread "main" java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to [Ljava.lang.String;
at ListToArrayTest.main(ListToArrayTest.java:23)


oboject的强转不是应该可以的吗?谢谢!



------解决方案--------------------
Java code
import java.util.ArrayList;

public class ListToArrayTest {
    public static void main(String args[]) {
        ArrayList<String> list = new ArrayList<String>();
        String group[];
        list.add("aaa");
        list.add("bbb");
        list.add("ccc");
        list.add("ddd");

        // 1
        // Object[] temp;
        // temp = list.toArray();
        // System.out.println(temp.length);
        // group=new String[temp.length];
        // for(int i=0;i<temp.length;i++){
        // group[i]=(String) temp[i];
        // }
        //
        // 2
        group = new String[list.size()];
        list.toArray(group);

        // Object to String

        System.out.println("group size is " + group.length);
        for(String s: list)
            System.out.println(s);
    }
}

------解决方案--------------------
Java code


    public static void main(String args[]) {
        ArrayList<String> list = new ArrayList<String>();
        String group[];
        list.add("aaa");
        list.add("bbb");
        list.add("ccc");
        list.add("ddd");
        String[] arr = new String[list.size()];
        list.toArray(arr);
       

        for (String s : arr)
        {
            System.out.println(s);
        }
      }

------解决方案--------------------
使用 public <T extends java/lang/Object> T[] toArray(T[]);

------解决方案--------------------
List list = new ArrayList();
list.add("aaa");
list.add("bbb");
list.add("ccc");
list.add("ddd");

int size = list.size();
String[] str = (String[])list.toArray(new String[size]);
------解决方案--------------------
list.toArray(group);
------解决方案--------------------
下面是这两种方法的区别,查源码和注释就明白了
Java code
 /**
     * Returns an array containing all of the elements in this list
     * in proper sequence (from first to last element).
     *
     * <p>The returned array will be "safe" in that no references to it are
     * maintained by this list.  (In other words, this method must allocate
     * a new array).  The caller is thus free to modify the returned array.
     *
     * <p>This method acts as bridge between array-based and collection-based
     * APIs.
     *
     * @return an array containing all of the elements in this list in
     *         proper sequence
     */
    public Object[] toArray() {
        return Arrays.copyOf(elementData, size);
    }

    /**
     * Returns an array containing all of the elements in this list in proper
     * sequence (from first to last element); the runtime type of the returned
     * array is that of the specified array.  If the list fits in the
     * specified array, it is returned therein.  Otherwise, a new array is
     * allocated with the runtime type of the specified array and the size of
     * this list.
     *
     * <p>If the list fits in the specified array with room to spare
     * (i.e., the array has more elements than the list), the element in
     * the array immediately following the end of the collection is set to
     * <tt>null</tt>.  (This is useful in determining the length of the
     * list <i>only</i> if the caller knows that the list does not contain
     * any null elements.)
     *
     * @param a the array into which the elements of the list are to
     *          be stored, if it is big enough; otherwise, a new array of the
     *          same runtime type is allocated for this purpose.
     * @return an array containing the elements of the list
     * @throws ArrayStoreException if the runtime type of the specified array
     *         is not a supertype of the runtime type of every element in
     *         this list
     * @throws NullPointerException if the specified array is null
     */
    public <T> T[] toArray(T[] a) {
        if (a.length < size)
            // Make a new array of a's runtime type, but my contents:
            return (T[]) Arrays.copyOf(elementData, size, a.getClass());
    System.arraycopy(elementData, 0, a, 0, size);
        if (a.length > size)
            a[size] = null;
        return a;
    }