Question

How is it possible that following code even compiles? As far as I can see the count function is called with two different types, yet compiler doesn’t complain and happily compiles this code.

public class Test {
        public static <T> int count(T[] x,T y){
                int count = 0;
                for(int i=0; i < x.length; i++){
                        if(x[i] == y) count ++;
                }
                return count;  
        }
        public static void main(String[] args) {
                Integer [] data = {1,2,3,1,4};
                String value = "1";
                int r =count(data,value);
                System.out.println( r + " - " + value);
        }
}

Answer 1

In this case the T is useless. You can change the signature to public static int count(Object[] x, Object y) without any effect on what arguments the compiler will let it accept. (You can see that the signature for Arrays.fill() uses that as the signature.)

If we consider the simpler case, where you just have arguments of type T, you can see that, since any instance of T is also an instance of its superclasses, T can always to be inferred to be its upper bound, and it will still accept the same argument types as before. Thus we can get rid of T and use its upper bound (in this case Object) instead.

Arrays in Java work the same way: arrays are covariant, which means that if S is a subclass of T, S[] is a subclass of T[]. So the same argument as above applies — if you just have arguments of type T and T[], T can be replaced by its upper bound.

(Note that this does not apply to generic types, which are not covariant or contravariant: List<S> is not a subtype of List<T>.)