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)

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;
    }


}

您可以创建如下所示的实用程序类，并使用它来抑制未检查的警告。

public class Objects {

    /**
     * Helps to avoid using {@code @SuppressWarnings({"unchecked"})} when casting to a generic type.
     */
    @SuppressWarnings({"unchecked"})
    public static <T> T uncheckedCast(Object obj) {
        return (T) obj;
    }
}

你可以这样使用它:

import static Objects.uncheckedCast;
...

HashMap<String, String> getItems(javax.servlet.http.HttpSession session) {
      return uncheckedCast(session.getAttribute("attributeKey"));
}

这里有更多关于这个问题的讨论: http://cleveralias.blogs.com/thought_spearmints/2006/01/suppresswarning.html

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);

快速猜测一下你是否发布了你的代码，可以肯定地说，但是你可能做了一些类似的事情

HashMap<String, Object> test = new HashMap();

当你需要做的时候，哪个会产生警告

HashMap<String, Object> test = new HashMap<String, Object>();

也许值得一看

Java编程语言中的泛型

如果你不熟悉需要做什么。

以这个为例，它比创建一个新的HashMap要快得多，如果它已经是一个HashMap，但仍然是安全的，因为每个元素都根据它的类型进行检查…

@SuppressWarnings("unchecked")
public static <K, V> HashMap<K, V> toHashMap(Object input, Class<K> key, Class<V> value) {
       assert input instanceof Map : input;

       for (Map.Entry<?, ?> e : ((HashMap<?, ?>) input).entrySet()) {
           assert key.isAssignableFrom(e.getKey().getClass()) : "Map contains invalid keys";
           assert value.isAssignableFrom(e.getValue().getClass()) : "Map contains invalid values";
       }

       if (input instanceof HashMap)
           return (HashMap<K, V>) input;
       return new HashMap<K, V>((Map<K, V>) input);
    }