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.
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。
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;
}
}
