The following topics are covered:
The section,
Generic Methods and Constructors, introduced you to type inference, which enables you to invoke a generic method as you would an ordinary method, without specifying a type between angle brackets. Consider the following example,
BoxDemo4
, which requires the
Box
example and Java SE 7 or later:
public class BoxDemo4 { public static <U> void addBox(U u, java.util.List<Box<U>> boxes) { Box<U> box = new Box<>(); box.add(u); boxes.add(box); } public static <U> void outputBoxes(java.util.List<Box<U>> boxes) { int counter = 0; for (Box<U> box: boxes) { U boxContents = box.get(); System.out.println( "Box #" + counter + " contains [" + boxContents.toString() + "]"); counter++; } } public static void main(String[] args) { java.util.ArrayList<Box<Integer>> listOfIntegerBoxes = new java.util.ArrayList<>(); BoxDemo4.<Integer>addBox(Integer.valueOf(10), listOfIntegerBoxes); BoxDemo4.addBox(Integer.valueOf(20), listOfIntegerBoxes); BoxDemo4.addBox(Integer.valueOf(30), listOfIntegerBoxes); BoxDemo4.outputBoxes(listOfIntegerBoxes); } }
The following is the output from this example:
Box #0 contains [10] Box #1 contains [20] Box #2 contains [30]
The generic method addBox
defines one type parameter named U
. Generally, the Java compiler can infer the type parameters of a generic method call. Consequently, in most cases, you do not have to specify them. For example, to call the generic method addBox
, you can specify the type parameter as follows:
BoxDemo4.<Integer>addBox(Integer.valueOf(10), listOfIntegerBoxes);
Alternatively, if you omit the type parameters, the Java compiler automatically infers (from the method's arguments) that the type parameter is Integer
:
BoxDemo4.addBox(Integer.valueOf(20), listOfIntegerBoxes);
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. This pair of angle brackets is informally called the diamond.
For example, consider the following variable declaration:
Map<String, List<String>> myMap = new HashMap<String, List<String>>();
In Java SE 7 and later, you can substitute the parameterized type of the constructor with an empty set of type parameters (<>
):
Map<String, List<String>> myMap = new HashMap<>();
Note that to take advantage of automatic type inference during generic class instantiation, you must specify the diamond. In the following example, the compiler generates an unchecked conversion warning because the HashMap()
constructor refers to the HashMap
raw type, not the Map<String, List<String>>
type:
Map<String, List<String>> myMap = new HashMap(); // unchecked conversion warning
Java SE 7 and later support limited type inference for generic instance creation; you can only use type inference if the parameterized type of the constructor is obvious from the context. For example, the following example does not compile:
List<String> list = new ArrayList<>(); list.add("A"); // The following statement should fail since addAll expects // Collection<? extends String> list.addAll(new ArrayList<>());
Note that the diamond often works in method calls; however, for greater clarity, it is suggested that you use the diamond primarily to initialize a variable where it is declared.
In comparison, the following example compiles:
// The following statements compile: List<? extends String> list2 = new ArrayList<>(); list.addAll(list2);
Note that constructors can be generic (in other words, declare their own formal type parameters) in both generic and non-generic classes. Consider the following example:
class MyClass<X> { <T> MyClass(T t) { // ... } }
Consider the following instantiation of the class MyClass
, which is valid in Java SE 7 and prior releases:
new MyClass<Integer>("")
This statement creates an instance of the parameterized type MyClass<Integer>
; the statement explicitly specifies the type Integer
for the formal type parameter, X
, of the generic class MyClass<X>
. Note that the constructor for this generic class contains a formal type parameter, T
. The compiler infers the type String
for the formal type parameter, T
, of the constructor of this generic class (because the actual parameter of this constructor is a String
object).
Compilers from releases prior to Java SE 7 are able to infer the actual type parameters of generic constructors, similar to generic methods. However, compilers in Java SE 7 and later can infer the actual type parameters of the generic class being instantiated if you use the diamond (<>
). Consider the following examples, which are valid for Java SE 7 and later:
MyClass<Integer> myObject = new MyClass<>("");
In this example, the compiler infers the type Integer
for the formal type parameter, X
, of the generic class MyClass<X>
. It infers the type String
for the formal type parameter, T
, of the constructor of this generic class.
MyClass<Integer> myObject = new <String> MyClass<>("");
In this example, the compiler infers the type Integer
for the formal type parameter, X
, of the generic class MyClass<X>
. The statement explicitly specifies the type String
for the formal type parameter, T
, of the constructor of this generic class.