Java接口(interface)抛出异常但接口(interface)实现不抛出异常？

Question

我阅读了这段代码，其中接口(interface)抛出异常，但实现它的类没有抛出或捕获异常，这是为什么呢？在java中它是合法的还是安全的？

import java.rmi.*;
public interface MyRemote extends Remote {
    public String sayHello() throws RemoteException;
}

import java.rmi.*;
import java.rmi.server.*;
public class MyRemoteImpl extends UnicastRemoteObject implements MyRemote{
    public String sayHello() {
        return "Server says, 'Hey'";
    }
    public MyRemoteImpl() throws RemoteException {}
    public static void main (String[] args) {
        try {
             MyRemote service = new MyRemoteImpl();
             Naming.rebind("RemoteHello", service);
        } catch(Exception ex)
        {
            ex.printStackTrace();
        }
    }
}

Answer 1

A general rule of implementing and extending is you can make your new class or interface "less restrictive" but not "more restrictive". If you think of the requirement to handle an exception as a restriction, an implementation that doesn't declare the exception is less restrictive. Anybody who codes to the interface will not have trouble with your class.

— 斯坦·詹姆斯

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作为 http://www.coderanch.com/t/399874/java/java/Methods-throwing-Exception-Interface 讨论的一部分