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.

当前回答

哇;我想我找到了自己问题的答案。我只是不确定这是否值得!:)

问题是演员没有检查。所以,你必须自己检查。不能只使用instanceof检查参数化类型,因为参数化类型信息在运行时不可用,在编译时已被擦除。

但是,您可以使用instanceof对散列中的每一项执行检查,这样就可以构造一个类型安全的新散列。你不会引起任何警告。

多亏了mmyers和Esko Luontola,我将最初在这里编写的代码参数化了,所以它可以被包装在某个实用程序类中,并用于任何参数化的HashMap。如果您想更好地理解它,并且对泛型不是很熟悉,我建议您查看这个答案的编辑历史。

public static <K, V> HashMap<K, V> castHash(HashMap input,
                                            Class<K> keyClass,
                                            Class<V> valueClass) {
  HashMap<K, V> output = new HashMap<K, V>();
  if (input == null)
      return output;
  for (Object key: input.keySet().toArray()) {
    if ((key == null) || (keyClass.isAssignableFrom(key.getClass()))) {
        Object value = input.get(key);
        if ((value == null) || (valueClass.isAssignableFrom(value.getClass()))) {
            K k = keyClass.cast(key);
            V v = valueClass.cast(value);
            output.put(k, v);
        } else {
            throw new AssertionError(
                "Cannot cast to HashMap<"+ keyClass.getSimpleName()
                +", "+ valueClass.getSimpleName() +">"
                +", value "+ value +" is not a "+ valueClass.getSimpleName()
            );
        }
    } else {
        throw new AssertionError(
            "Cannot cast to HashMap<"+ keyClass.getSimpleName()
            +", "+ valueClass.getSimpleName() +">"
            +", key "+ key +" is not a " + keyClass.getSimpleName()
        );
    }
  }
  return output;
}

工作量很大,报酬可能很少……我不确定我是否会用它。我很感激任何关于人们是否认为它值得的评论。此外,我很感激改进建议:除了抛出AssertionErrors之外,还有什么更好的事情我可以做吗?有更好的东西让我扔吗?我应该让它成为一个受控异常吗?

其他回答

在这种特殊情况下,我不会将Maps直接存储到HttpSession中,而是将我自己的类的实例存储到HttpSession中,而该类的实例又包含一个Map(类的实现细节)。然后可以确定映射中的元素是正确的类型。

但是如果你想检查Map的内容类型是否正确,你可以使用这样的代码:

public static void main(String[] args) {
    Map<String, Integer> map = new HashMap<String, Integer>();
    map.put("a", 1);
    map.put("b", 2);
    Object obj = map;

    Map<String, Integer> ok = safeCastMap(obj, String.class, Integer.class);
    Map<String, String> error = safeCastMap(obj, String.class, String.class);
}

@SuppressWarnings({"unchecked"})
public static <K, V> Map<K, V> safeCastMap(Object map, Class<K> keyType, Class<V> valueType) {
    checkMap(map);
    checkMapContents(keyType, valueType, (Map<?, ?>) map);
    return (Map<K, V>) map;
}

private static void checkMap(Object map) {
    checkType(Map.class, map);
}

private static <K, V> void checkMapContents(Class<K> keyType, Class<V> valueType, Map<?, ?> map) {
    for (Map.Entry<?, ?> entry : map.entrySet()) {
        checkType(keyType, entry.getKey());
        checkType(valueType, entry.getValue());
    }
}

private static <K> void checkType(Class<K> expectedType, Object obj) {
    if (!expectedType.isInstance(obj)) {
        throw new IllegalArgumentException("Expected " + expectedType + " but was " + obj.getClass() + ": " + obj);
    }
}

如果我不得不使用一个不支持泛型的API ..我尝试用尽可能少的行将这些调用隔离在包装器例程中。然后我使用SuppressWarnings注释并同时添加类型安全类型强制转换。

这只是个人的喜好,想让事情尽可能的整洁。

在Eclipse首选项中,转到Java->编译器->错误/警告->泛型类型,并选中“忽略不可避免的泛型类型问题”复选框。

这满足了问题的意图,即。

我希望避免Eclipse警告……

如果不是精神。

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);

下面是重写equals()操作时的一种处理方法。

public abstract class Section<T extends Section> extends Element<Section<T>> {
    Object attr1;

    /**
    * Compare one section object to another.
    *
    * @param obj the object being compared with this section object
    * @return true if this section and the other section are of the same
    * sub-class of section and their component fields are the same, false
    * otherwise
    */       
    @Override
    public boolean equals(Object obj) {
        if (obj == null) {
            // this exists, but obj doesn't, so they can't be equal!
            return false;
        }

        // prepare to cast...
        Section<?> other;

        if (getClass() != obj.getClass()) {
            // looks like we're comparing apples to oranges
            return false;
        } else {
            // it must be safe to make that cast!
            other = (Section<?>) obj;
        }

        // and then I compare attributes between this and other
        return this.attr1.equals(other.attr1);
    }
}

这似乎在Java 8中工作(甚至用-Xlint:unchecked编译)