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

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。

其他回答

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;
    }


}

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);
    }
}

不幸的是,这里没有很好的选择。记住,所有这些的目标都是保持类型安全。“Java泛型”提供了一种处理非泛型遗留库的解决方案,在8.2节中有一种特别的方法叫做“空循环技术”。基本上,使不安全施放,并压制警告。然后像这样循环地图:

@SuppressWarnings("unchecked")
Map<String, Number> map = getMap();
for (String s : map.keySet());
for (Number n : map.values());

如果遇到意外类型,您将得到一个运行时ClassCastException,但至少它将发生在问题的根源附近。

当然,最明显的答案是不要进行未经检查的强制转换。

如果绝对有必要,那么至少尝试限制@SuppressWarnings注释的范围。根据它的Javadocs,它可以访问局部变量;这样,它甚至不会影响整个方法。

例子:

@SuppressWarnings("unchecked")
Map<String, String> myMap = (Map<String, String>) deserializeMap();

没有办法确定Map是否真的应该有通用参数<String, String>。您必须事先知道参数应该是什么(否则当您得到ClassCastException时就会知道)。这就是代码生成警告的原因,因为编译器不可能知道是否安全。

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。