看代碼看到
public Runnable r = new Runnable() { @Override public void run() { ... } }
接口不能new ,不過可以生成一個匿名類,省略了寫一個具體類實現接口的開銷。
public class Main { public static void main(String[] args) { String a=new CustomerImpl().sayHello("hongda"); System.out.println(a); String b=new Customer(){ public String sayHello(String name){ return "Hello2 "+name; } }.sayHello("hongdada"); System.out.println(b); } } interface Customer { public String sayHello(String name); } class CustomerImpl implements Customer { @Override public String sayHello(String name) { return "Hello, " + name; } }
Hello, hongda
Hello2 hongdada
如果接口內有多個方法呢?
public class Main { public static void main(String[] args) { String a=new CustomerImpl().sayHello("hongda"); System.out.println(a); String b=new Customer(){ public String sayHello(String name){ return "Hello2 "+name; } }.sayHello("hongdada"); System.out.println(b); } } interface Customer { public String sayHello(String name); public int Add(int a ,int b); } class CustomerImpl implements Customer { @Override public String sayHello(String name) { return "Hello, " + name; } @Override public int Add(int a ,int b){ return a+b; } }
Error:(8, 32) java: <匿名com.company.Main$1>不是抽象的, 並且未覆蓋com.company.Customer中的抽象方法Add(int,int)
匿名類內部實現接口全部方法:
public class Main { public static void main(String[] args) { String a=new CustomerImpl().sayHello("hongda"); System.out.println(a); String b=new Customer(){ public String sayHello(String name){ return "Hello2 "+name; } public int Add(int a ,int b){ return a+b; } }.sayHello("hongdada"); System.out.println(b); } } interface Customer { public String sayHello(String name); public int Add(int a ,int b); } class CustomerImpl implements Customer { @Override public String sayHello(String name) { return "Hello, " + name; } @Override public int Add(int a ,int b){ return a+b; } }
Hello, hongda
Hello2 hongdada
這種實現接口方式其實就是一個實現一個繼承接口的匿名類。
