在这里,我试图将地图值转换为字符串数组但我得到了
错误
ERROR/AndroidRuntime(23588): Caused by: java.lang.ClassCastException: [Ljava.lang.Object;
代码
Map<String,String> contactNumber = new HashMap<String,String>();
String results [] = (String[]) contactNumber.values().toArray();
答案 0 :(得分:24)
您应该使用其他toArray(T[] a)方法。
String[] result = contactNumber.values().toArray(new String[0]);
答案 1 :(得分:7)
你不能像这样执行演员表。相反,请调用另一个toArray方法:
String[] result = contactNumber.values().toArray(new String[0]);
答案 2 :(得分:4)
您正在调用Collection接口上的toArray()方法,该方法返回一个Object数组。切换到toArray(T [] a)可以帮助您解决问题。 javadocs应该有助于进一步澄清答案。
/**
* Returns an array containing all of the elements in this collection.
* If this collection makes any guarantees as to what order its elements
* are returned by its iterator, this method must return the elements in
* the same order.
*
* <p>The returned array will be "safe" in that no references to it are
* maintained by this collection. (In other words, this method must
* allocate a new array even if this collection is backed by an 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 collection
*/
Object[] toArray();
/**
* Returns an array containing all of the elements in this collection;
* the runtime type of the returned array is that of the specified array.
* If the collection 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 collection.
*
* <p>If this collection fits in the specified array with room to spare
* (i.e., the array has more elements than this collection), 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 this
* collection <i>only</i> if the caller knows that this collection does
* not contain any <tt>null</tt> elements.)
*
* <p>If this collection makes any guarantees as to what order its elements
* are returned by its iterator, this method must return the elements in
* the same order.
*
* <p>Like the {@link #toArray()} method, this method acts as bridge between
* array-based and collection-based APIs. Further, this method allows
* precise control over the runtime type of the output array, and may,
* under certain circumstances, be used to save allocation costs.
*
* <p>Suppose <tt>x</tt> is a collection known to contain only strings.
* The following code can be used to dump the collection into a newly
* allocated array of <tt>String</tt>:
*
* <pre>
* String[] y = x.toArray(new String[0]);</pre>
*
* Note that <tt>toArray(new Object[0])</tt> is identical in function to
* <tt>toArray()</tt>.
*
* @param a the array into which the elements of this collection 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 all of the elements in this collection
* @throws ArrayStoreException if the runtime type of the specified array
* is not a supertype of the runtime type of every element in
* this collection
* @throws NullPointerException if the specified array is null
*/
<T> T[] toArray(T[] a);