As you may know, one of new features of upcoming Java 7 will be the diamond operator. Purpose of the diamond operator is to simplify instantiation of generic classes. For example, instead of
List<Integer> p = new ArrayList<Integer>();
with the diamond operator we can write only
List<Integer> p = new ArrayList<>();
and let compiler infer the value of type argument. Nice simplification. But do we really need to write <>? Isn't new ArrayList() enough? In this article, I will describe the arguments of the <> proponents and explain why I think that these arguments are not very strong. However, I also describe arguments why we need <>.
In Java 1.4, we had raw types only:
List p = new ArrayList();
Java 5 introduced generics:
List<Integer> p = new ArrayList<Integer>();
Many types in Java API were generified and even though we can still use generic types as raw types, there is no reason for this in Java 5 or newer. When generics were introduced, raw types were allowed for backward compatibility so that we could gradually and smoothly adopt generics. For example, code in Java 1.4 can be combined with new generic code because raw and generic types are allowed together. This is also expressed in the JLS (4.8 Raw Types):
"The use of raw types is allowed only as a concession to compatibility of legacy code. The use of raw types in code written after the introduction of genericity into the Java programming language is strongly discouraged. It is possible that future versions of the Java programming language will disallow the use of raw types."
Now let's go back to the diamond operator and ask again: "Do we really need <>?". The proponents of the <> syntax say that we need <> to preserve backward compatibility. Let's look at an example from the coin-dev conference:
class Foo<X> {
Foo(X x) { }
Foo<X> get(X x) { return this; }
}
class Test {
void test() {
Foo<?> f1 = new Foo(1).get(""); //ok - can pass String where Object is expected
Foo<?> f2 = new Foo<>(1).get(""); //fail - cannot pass String where Integer is expected
}
}
This shows the difference between new Foo(1) and new Foo<>(1). Clearly, these two are different and if we changed the semantics of new Foo(1), it would break backward compatibility. But wait. Backward compatibility with what? Isn't line
Foo<?> f1 = new Foo(1).get("");
a little suspicious? It uses generic type in the left part and raw type in the right part. Although it is legal, it is probably either omission or malpractice. And its legality is probably only a side effect of "a concession to compatibility of legacy code".
Let's go further and look at another example from the coin-dev conference. It shows the difference between raw type and parameterized type with the diamond:
public class X<T> {
public X(T t) { }
public T get() { return null; }
public static int f(String s) { return 1; }
public static int f(Object o) { return 2; }
public static void main(String[] args) {
System.out.println(f(new X<>("").get()));
System.out.println(f(new X("").get()));
}
}
Let's play with the code a bit. Let's assume that there was a library with the X class:
public class X {
public X(Object o) { }
public Object get() { return null; }
}
and some code that compiled against this library:
public class Client {
static int f(String s) { return 1; }
static int f(Object o) { return 2; }
public static void main(String[] args) {
System.out.println(f(new X("").get()));
}
}
Then, the library was generified:
public class X<T> {
public X(T t) { }
public T get() { return null; }
}
and we compiled the client project against the generified version. Now, if we changed the semantics of new X("") to new X<String>("") (or new X<>("") with the diamond syntax), the code would behave differently. So, the answer to the title question is 'yes'. If we want to stay backward compatible, we need <> and we cannot put new X("") semantically equal to new X<>("").
Other questions are how long can Java evolve and remain compatible with concessions to compatibility and whether newcomers to Java will appreciate this.
