Summary of the usage of generics in Java _java

The example in this article summarizes the usage of generics in Java. Share to everyone for your reference. Specifically as follows:

1 Basic use

Public interface List<e> {
 void Add (E);
 Iterator<e> iterator ();
}

2 Generics and subclasses

The child is a subclass of Parent,list<child> but not a list<parent> subclass.
therefore:list<object> List = new arraylist<string> () 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, converting Object to String will fail.

3 wildcards

Because of the 2 reason, the following implementations are not available for the output of the wildcard collection.

void Printcollection (collection<object> c) {for
 (Object o:c) {
 //do Something
 }
}

Therefore, wildcard characters are required:

void Printcollection (collection<?> c) {for
 (Object o:c) {//1
 /do Something
 }
}//OK

The type is unknown, but any objects are object, so 1 of the example above is correct. But the following example is wrong:

void Add (collection<. extends myclass> c) {
 C.add (new MyClass ());//Wrong
}//OK

The reason is clear, too? Extends MyClass description type is a subclass of MyClass, but does not know the specific type

4. Generic method

The example above can be implemented as:

<T> Add (collection<t> C, T t) {
 c.add (t);
}

The compiler will help with the conversion of the type in the premise of guaranteeing semantics.

5. Comparison of generic run-time

list<string> L1 = new arraylist<string> ();
list<integer> L2 = new arraylist<integer> ();
System.out.println (l1.getclass () = = L2.getclass ()); True

Because the generic class runs as always.

6 generic Array (may cause type insecurity)

Copy Code code as follows:

list<string>[] LSA = new arraylist<string>[10]; Error

If you can, it may cause 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 help you with your Java programming.