在Java类中如何隐藏不受支持的接口方法？
实际上，一旦实现接口，就不能为其所有方法提供实现，或者使类抽象化。没有实现就无法跳过接口的方法（除非它们是默认方法）。但是，如果尝试跳过接口的实现方法，则会生成编译时错误。
示例
interface MyInterface{
public static int num = 100;
public void sample();
public void getDetails();
public void setNumber(int num);
public void setString(String data);
}
public class InterfaceExample implements MyInterface{
public static int num = 10000;
public void sample() {
System.out.println("这是sample方法的实现");
}
public static void main(String args[]) {
InterfaceExample obj = new InterfaceExample();
obj.sample();
}
}输出结果
编译时错误
InterfaceExample.java:8: error: InterfaceExample is not abstract and does not override abstract method setString(String) in MyInterface
public class InterfaceExample implements MyInterface{
^
1 error但是，您可以实现这些不需要的/不受支持的方法，并从中抛出异常，例如UnsupportedOperationException或IllegalStateException。
示例
interface MyInterface{
public void sample();
public void getDetails();
public void setNumber(int num);
public void setString(String data);
}
public class InterfaceExample implements MyInterface{
public void getDetails() {
try {
throw new UnsupportedOperationException();
}
catch(UnsupportedOperationException ex) {
System.out.println("Method not supported");
}
}
public void setNumber(int num) {
try {
throw new UnsupportedOperationException();
}
catch(UnsupportedOperationException ex) {
System.out.println("Method not supported");
}
}
public void setString(String data) {
try {
throw new UnsupportedOperationException();
}
catch(UnsupportedOperationException ex) {
System.out.println("Method not supported");
}
}
public void sample() {
System.out.println("这是sample方法的实现");
}
public static void main(String args[]) {
InterfaceExample obj = new InterfaceExample();
obj.sample();
obj.getDetails();
obj.setNumber(21);
obj.setString("data");
}
}输出结果
这是sample方法的实现 Method not supported Method not supported Method not supported