Java – How to address unchecked cast warnings

genericsjavawarnings

Eclipse is giving me a warning of the following form:

Type safety: Unchecked cast from Object to HashMap

This is from a call to an API that I have no control over which returns Object:

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

I'd like to avoid Eclipse warnings, if possible, since theoretically they indicate at least a potential code problem. I haven't found a good way to eliminate this one yet, though. I can extract the single line involved out to a method by itself and add @SuppressWarnings("unchecked") to that method, thus limiting the impact of having a block of code where I ignore warnings. Any better options? I don't want to turn these warnings off in Eclipse.

Before I came to the code, it was simpler, but still provoked warnings:

HashMap getItems(javax.servlet.http.HttpSession session) {
  HashMap theHash = (HashMap)session.getAttribute("attributeKey");
  return theHash;
}

Problem was elsewhere when you tried to use the hash you'd get warnings:

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.

Best Solution

The obvious answer, of course, is not to do the unchecked cast.

If it's absolutely necessary, then at least try to limit the scope of the @SuppressWarnings annotation. According to its Javadocs, it can go on local variables; this way, it doesn't even affect the entire method.

Example:

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

There is no way to determine whether the Map really should have the generic parameters <String, String>. You must know beforehand what the parameters should be (or you'll find out when you get a ClassCastException). This is why the code generates a warning, because the compiler can't possibly know whether is safe.