java - How do I address unchecked cast warnings?

Question

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.

Answer 1

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.