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

在Android Studio中,如果你想禁用检查,你可以使用:

//noinspection unchecked
Map<String, String> myMap = (Map<String, String>) deserializeMap();

其他回答

的对象。Esko Luontola上面回答的未检查的实用函数是避免程序混乱的好方法。

如果您不希望在整个方法上使用SuppressWarnings, Java会强制您将其放在本地方法上。如果你需要对一个成员进行强制转换,可能会导致这样的代码:

@SuppressWarnings("unchecked")
Vector<String> watchedSymbolsClone = (Vector<String>) watchedSymbols.clone();
this.watchedSymbols = watchedSymbolsClone;

使用这个实用程序要干净得多,而且你所做的事情仍然很明显:

this.watchedSymbols = Objects.uncheckedCast(watchedSymbols.clone());

注意: 我觉得有必要补充一下,有时候警告真的意味着你做错了什么,比如:

ArrayList<Integer> intList = new ArrayList<Integer>();
intList.add(1);
Object intListObject = intList; 

 // this line gives an unchecked warning - but no runtime error
ArrayList<String> stringList  = (ArrayList<String>) intListObject;
System.out.println(stringList.get(0)); // cast exception will be given here

编译器告诉您的是,在运行时不会检查此强制转换,因此在尝试访问泛型容器中的数据之前不会引发运行时错误。

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);

在Android Studio中,如果你想禁用检查,你可以使用:

//noinspection unchecked
Map<String, String> myMap = (Map<String, String>) deserializeMap();

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?

Two ways, one which avoids the tag completely, the other using a naughty but nice utility method. The problem is pre-genericised Collections... I believe the rule of thumb is: "cast objects one thing at a time" - what this means when trying to use raw classes in a genericised world is that because you don't know what is in this Map<?, ?> (and indeed the JVM might even find that it isn't even a Map!), it obvious when you think about it that you can't cast it. If you had a Map<String, ?> map2 then HashSet<String> keys = (HashSet<String>)map2.keySet() does not give you a warning, despite this being an "act of faith" for the compiler (because it might turn out to be a TreeSet)... but it is only a single act of faith. PS to the objection that iterating as in my first way "is boring" and "takes time", the answer is "no pain no gain": a genericised collection is guaranteed to contain Map.Entry<String, String>s, and nothing else. You have to pay for this guarantee. When using generics systematically this payment, beautifully, takes the form of coding compliance, not machine time! One school of thought might say that you should set Eclipse's settings to make such unchecked casts errors, rather than warnings. In that case you would have to use my first way.

package scratchpad;

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Vector;

public class YellowMouse {

    // First way

    Map<String, String> getHashMapStudiouslyAvoidingSuppressTag(HttpSession session) {
      Map<?, ?> theHash = (Map<?, ?>)session.getAttribute("attributeKey");

      Map<String, String> yellowMouse = new HashMap<String, String>();
      for( Map.Entry<?, ?> entry : theHash.entrySet() ){
        yellowMouse.put( (String)entry.getKey(), (String)entry.getValue() );
      }

      return yellowMouse;
    }


    // Second way

    Map<String, String> getHashMapUsingNaughtyButNiceUtilityMethod(HttpSession session) {
      return uncheckedCast( session.getAttribute("attributeKey") );
    }


    // NB this is a utility method which should be kept in your utility library. If you do that it will
    // be the *only* time in your entire life that you will have to use this particular tag!!

    @SuppressWarnings({ "unchecked" })
    public static synchronized <T> T uncheckedCast(Object obj) {
        return (T) obj;
    }


}