Eclipse给我一个如下形式的警告:
类型安全:未检查从对象转换到HashMap
这是从一个API调用,我无法控制返回对象:
HashMap<String, String> getItems(javax.servlet.http.HttpSession session) {
HashMap<String, String> theHash = (HashMap<String, String>)session.getAttribute("attributeKey");
return theHash;
}
如果可能的话,我希望避免使用Eclipse警告,因为理论上它们至少表明存在潜在的代码问题。不过,我还没有找到消除这个问题的好方法。我可以将所涉及的单行单独提取到一个方法中,并向该方法添加@SuppressWarnings(“unchecked”),从而限制忽略警告的代码块的影响。有更好的选择吗?我不想在Eclipse中关闭这些警告。
在我接触代码之前,它更简单,但仍然引起了警告:
HashMap getItems(javax.servlet.http.HttpSession session) {
HashMap theHash = (HashMap)session.getAttribute("attributeKey");
return theHash;
}
问题是在其他地方,当你试图使用散列时,你会得到警告:
HashMap items = getItems(session);
items.put("this", "that");
Type safety: The method put(Object, Object) belongs to the raw type HashMap. References to generic type HashMap<K,V> should be parameterized.
计算机科学中的几乎每一个问题都可以通过添加一层间接方法来解决。
因此,引入一个非泛型对象,它的级别高于Map。如果没有上下文,它看起来不会很有说服力,但无论如何:
public final class Items implements java.io.Serializable {
private static final long serialVersionUID = 1L;
private Map<String,String> map;
public Items(Map<String,String> map) {
this.map = New.immutableMap(map);
}
public Map<String,String> getMap() {
return map;
}
@Override public String toString() {
return map.toString();
}
}
public final class New {
public static <K,V> Map<K,V> immutableMap(
Map<? extends K, ? extends V> original
) {
// ... optimise as you wish...
return Collections.unmodifiableMap(
new HashMap<String,String>(original)
);
}
}
static Map<String, String> getItems(HttpSession session) {
Items items = (Items)
session.getAttribute("attributeKey");
return items.getMap();
}
*除了过多的间接层次。
下面是一个简短的示例,通过使用其他回答中提到的两种策略来避免“unchecked cast”警告。
Pass down the Class of the type of interest as a parameter at runtime (Class<T> inputElementClazz). Then you can use: inputElementClazz.cast(anyObject);
For type casting of a Collection, use the wildcard ? instead of a generic type T to acknowledge that you indeed do not know what kind of objects to expect from the legacy code (Collection<?> unknownTypeCollection). After all, this is what the "unchecked cast" warning wants to tell us: We cannot be sure that we get a Collection<T>, so the honest thing to do is to use a Collection<?>. If absolutely needed, a collection of a known type can still be built (Collection<T> knownTypeCollection).
下面示例中的遗留代码接口在StructuredViewer中有一个属性“input”(StructuredViewer是一个树或表小部件,“input”是它背后的数据模型)。这个“输入”可以是任何类型的Java集合。
public void dragFinished(StructuredViewer structuredViewer, Class<T> inputElementClazz) {
IStructuredSelection selection = (IStructuredSelection) structuredViewer.getSelection();
// legacy code returns an Object from getFirstElement,
// the developer knows/hopes it is of type inputElementClazz, but the compiler cannot know
T firstElement = inputElementClazz.cast(selection.getFirstElement());
// legacy code returns an object from getInput, so we deal with it as a Collection<?>
Collection<?> unknownTypeCollection = (Collection<?>) structuredViewer.getInput();
// for some operations we do not even need a collection with known types
unknownTypeCollection.remove(firstElement);
// nothing prevents us from building a Collection of a known type, should we really need one
Collection<T> knownTypeCollection = new ArrayList<T>();
for (Object object : unknownTypeCollection) {
T aT = inputElementClazz.cast(object);
knownTypeCollection.add(aT);
System.out.println(aT.getClass());
}
structuredViewer.refresh();
}
当然,如果我们使用错误的数据类型的遗留代码(例如,如果我们将一个数组设置为StructuredViewer的“输入”而不是Java Collection),上面的代码就会给出运行时错误。
调用方法的例子:
dragFinishedStrategy.dragFinished(viewer, Product.class);