Why can a Double be added to a List of Integers using reflection

Question

Why does this code run without any exceptions?

public static void main(String args[]) {     List<Integer> a = new ArrayList<Integer>();     try {          a.getClass()             .getMethod("add", Object.class)             .invoke(a, new Double(0.55555));      } catch (Exception e) {         e.printStackTrace();     }      System.out.println(a.get(0)); } 

Answer 1

Generics are a compile-time thing. At runtime, a regular ArrayList, without any additional check, is used. Since you're bypassing the safety checks by using reflection to add elements to your list, nothing can prevent a Double from being stored inside your List<Integer>. Just like if you did

List<Integer> list = new ArrayList<Integer>(); List rawList = list; rawList.add(new Double(2.5)); 

If you want your list to implement type checks at runtime, then use

List<Integer> checkedList = Collections.checkedList(list, Integer.class); 

Answer 2

Because of the type erasure - there are no runtime checks for the generics, during compilation type parameters are removed: Java generics - type erasure - when and what happens.

You may be surprised, but you don't need to use reflection to add a Double to a List<Integer>:

List<Integer> a = new ArrayList<Integer>(); ((List)a).add(new Double(0.555));