Summary of usage of generics in Java

The examples in this article summarize the usage of generics in Java. Share it with everyone for your reference. The details are as follows:

1 Basic use

public interface List<E> {
 void add(E);
 Iterator<E> iterator();
}

2 Generics and subclasses

Child is a subclass of Parent, but Listad7a74ca7e6896ff8ac9eb8e25d760c6 is not A subclass of List294ec52526076f8c76b491d15e4edc9c.
So: Lista87fdacec66f0909fc0757c19f2d2b1d list = new ArrayListf7e83be87db5cd2d9a8a0b8117b38cd4() is wrong.
If the above is correct, then:

List<String> ls = new ArrayList<String>(); //1
List<Object> lo = ls; //2
lo.add(new Object()); // 3
String s = ls.get(0); // 4，将object转换为string将失败。

3 wildcards

Because of reason 2, the following implementation cannot be used to unify the output of the collection.

void printCollection(Collection<Object> c) {
 for (Object o: c) {
 // do something
 }
}

Therefore, the wildcard ? is needed:

void printCollection(Collection<?> c) {
 for (Object o: c) { // 1
 // do something
 }
} // ok

The ? here indicates that the type is unknown, but any object is Object, so 1 in the above example is correct. But the following example is wrong:

void add(Collection<? extends MyClass> c) {
 c.add(new MyClass()); // wrong
} // ok

The reason is also very clear, ? extends MyClass indicates that the type is a subclass of MyClass, but the specific type is not known

4. Generic method

The above example can be implemented as:

<T> add(Collection<T> c, T t) {
 c.add(t);
}

The compiler will help with type conversion while ensuring semantics.

5. Comparison of generic runtime

List<String> l1 = new ArrayList<String>();
List<Integer> l2 = new ArrayList<Integer>();
System.out.println(l1.getClass() == l2.getClass()); // true

Because the runtime of generic classes is the same.

6 Generic array (may lead to type insecurity)

List<String>[] lsa = new ArrayList<String>[10]; // error

If possible, may lead to type insecurity. Such as:

Object o = lsa;
Object []oa = (Object[])o;
List<Integer> li = new ArrayList<Integer>();
li.add(new Integer(3));
oa[1] = li;
String s = lsa[1].get(0); // runtime error

I hope this article will be helpful to everyone’s java programming.

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