Effective Java 服務提供者框架

2019-11-14 23:29:20

字體：大中小

來源：轉載

供稿：網友

Effective java 服務提供者框架

注：今天看《Effective Java 中文版第2版》時，發現在靜態工廠方法中提到的服務提供者框架的簡單實現代碼不夠完整，所以自己寫了一份完整代碼附帶一些解釋，希望對大家有所幫助。

Service.java

// 服務接口public interface Service {    public abstract void serve();}

PRovider.java

// 提供者接口public interface Provider {    Service newService();}

ServiceManager.java

import java.util.Map;import java.util.concurrent.ConcurrentHashMap;public class ServiceManager {    private ServiceManager() {}        // 把服務接口的名字映射到Services中    private static final Map<String, Provider> providers =             new ConcurrentHashMap<String, Provider>();    public static final String DEFAULT_PROVIDER_NAME = "<def>";        // 提供者注冊API(默認)    public static void registerDefaultProvider(Provider p) {        registerProvider(DEFAULT_PROVIDER_NAME, p);    }

    // 提供者注冊API    public static void registerProvider(String name, Provider p) {        providers.put(name, p);    }        // 服務訪問API(默認)    public static Service newInstance() {        return newInstance(DEFAULT_PROVIDER_NAME);    }    // 服務訪問API

    public static Service newInstance(String name) {                Provider p = providers.get(name);        if(p == null) {            throw new IllegalArgumentException("No provider registered with name: " + name);        }        return p.newService();    }}

EntityProvider.java

public class EntityProvider implements Provider {    static {        ServiceManager.registerProvider("EntityService", new EntityProvider());    }        @Override    public Service newService() {        return new ServiceImpl();    }        class ServiceImpl implements Service {        @Override        public void serve() {            System.out.println("開始服務...");        }    }}

Test.java

public class Test {    public static void main(String[] args) throws ClassNotFoundException {                Class.forName("EntityProvider");                 Service s = ServiceManager.newInstance("EntityService");        s.serve();