Question

My code is:

import java.util.*;

public class A {
public static void main(String[] args){
    List<String> list = new ArrayList();
    list.add("1"); //ok   line 1
    list.add(1); //error  line 2
}

When I run this code Java gives me an error and I know why, but even when I only use line 1 the compiler warns me. Why do I get this warning? I don’t understand what is the difference between my first example and this code:

import java.util.*;

public class A {
    public static void main(String[] args){
        List<String> list = new ArrayList<String>(); // <-- notice the second <String>
        list.add("1"); //ok   line 1
        list.add(1); //error  line 2
    }
}

Answer 1

When I run this code Java gives me an error and I know why, but even when I only use line 1 the compiler warns me. Why do I get this warning?

When you use your code, compiler will give you a warning like

List<String> list = new ArrayList();

ArrayList is a raw type. References to generic type ArrayList should be parameterized

It says that you have to provide a parameterized type for your ArrayList as well.

List<String> list = new ArrayList<String>();

EDIT:

As sugguested by @newacct: if you are using Java 7, you could use empty angular brackets (<>) when instantiating your Collection like:

List<String> list = new ArrayList<>();

From the Java 7 documentation:

You can replace the type arguments required to invoke the constructor of a generic class with an empty set of type parameters (<>) as long as the compiler can infer the type arguments from the context.