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

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。

其他回答

在强制转换之前检查一下。

Object someObject = session.getAttribute("attributeKey");
if(someObject instanceof HashMap)
HashMap<String, String> theHash = (HashMap<String, String>)someObject;  

对于任何提问的人来说,收到你不确定类型的对象是很常见的。大量遗留的“SOA”实现传递各种对象,您不应该总是信任这些对象。(恐怖!)

EDIT修改了示例代码一次,以匹配海报的更新,在一些评论之后,我看到instanceof不能很好地使用泛型。然而,更改检查以验证外部对象似乎可以很好地使用命令行编译器。修订的例子现在发布。

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。

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;
    }


}

我可能误解了这个问题(一个示例和几行代码就好了),但是为什么您不总是使用合适的接口(和Java5+)呢?我看不出为什么你想要强制转换到HashMap而不是Map<KeyType,ValueType>。事实上,我想不出有什么理由将变量的类型设置为HashMap而不是Map。

为什么源是一个对象?是遗留集合的参数类型吗?如果是,请使用泛型并指定所需的类型。