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What is the meaning of the type safety warning in certain Java generics casts?

By : acacio
Date : March 29 2020, 07:55 AM
To fix the issue you can do This warning is there because Java is not actually storing type information at run-time in an object that uses generics. Thus, if object is actually a List, there will be no ClassCastException at run-time except until an item is accessed from the list that doesn't match the generic type defined in the variable.
This can cause further complications if items are added to the list, with this incorrect generic type information. Any code still holding a reference to the list but with the correct generic type information will now have an inconsistent list.
code :
List<?> list = (List<?>) object;

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Java Generics Type Safety warning with recursive Hashmap

By : oosbuey Kang
Date : March 29 2020, 07:55 AM
To fix the issue you can do This is possible using a generic method with a recursive type variable. Try the following:
code :
public <T extends Map<String, T>> void foo(String filename, T map) {
    //some stuff here
    for (Map.Entry<String, T> entry : map.entrySet())  {
        foo(entry.getKey(), entry.getValue());
public class Node {
    private Map<String, Node> children;

    // accessor methods to retrieve children ...

Why do I get a Type safety warning here? (generics)

By : Oscar Morocho
Date : March 29 2020, 07:55 AM
I wish this help you I lately wanted to learn more about generic methods and created several examples, but the following type safety is not clear to me. Here an example:
code :
class AbstractRepository

  Set<? extends SuperObject> getObject();

class Repository extends AbstractRepository

  Set<    ConcreteObject   > getObject()

Eclipse warning: Type safety (Java Generics)

By : Badar Ali
Date : March 29 2020, 07:55 AM
should help you out The signature of that class is HibernateCallback and defines a method T doInHibernate(Session) but you don't supply the type parameter T -- that's what the compiler is complaining about: It's not sure you're your code is actually resulting in a List that fits into your result variable.
You are correct in that adding @SuppressWarnings isn't a good idea (it doesn't increase actually type safety), try this instead:
code :
List<Book> result = hibernateTemplate.execute(new HibernateCallback<List<Book>>() {
    public List<Book> doInHibernate(Session session) throws HibernateException, SQLException {
        Query query = session.createQuery("SELECT DISTINCT b FROM Book as b LEFT JOIN FETCH b.authors");

        List list = query.list();

        return list;

JSON and Generics in Java - Type safety warning

By : David
Date : March 29 2020, 07:55 AM
I hope this helps you . What is your JSONObject, does it inherit from HashMap? If does, the warn probably means that your should declare the JSONObject instance as follows:
code :
JSONObject<String,Object> obj=new JSONObject<String,Object>();
public class JSONObject extends HashMap
public class JSONObject<K,V> extends HashMap<K,V>
JSONObject<String,Object> obj=new JSONObject<String,Object>();

Encountering type safety warning when using Generics

By : Luryson Leou
Date : March 29 2020, 07:55 AM
this will help I have a class that looks like this.
code :
public <K extends ClassB> K someMethod(T t) {
    return null;
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