public class OutterTest
{
public static void main(String[] args)
{
Outter out = new Outter();
out.show(11);
}
}
class Outter
{
public void show(final int a)
{
class Inner
{
public void show1()
{
System.out.println(a);
}
}
new Inner().show1();
}
}
为什么在方法内的类只能访问带final的局部变量呢?作者: 吴琼 时间: 2012-6-27 11:18
原因如下:
abstract class ABSClass{
public abstract void print();
}
public class Test2{
public static void test(final String s){//一旦参数在匿名类内部使用,则必须是final
ABSClass c=new ABSClass(){
public void print(){
System.out.println(s);
}
};
c.print();
}
public static void main(String[] args){
test("Hello World!");
}
}
public class MyClass {
public MyClass() {
final int finalValue = 10;
int not$Final = 20;
MyInterface myInterface = new MyInterface() {
public void functionWithoutPara() {
//compile Error
//System.out.println(noFinal);
System.out.println(finalValue);
}
public void functionWithPara(int num) {
System.out.println("The parameter " + num
+ " has been passed by the method");
}
public class MyClass {
public MyClass() {
final int finalValue = 10;
int not$Final = 20;
MyInterface myInterface = new MyInterface() {
public void functionWithoutPara() {
//compile Error
//System.out.println(noFinal);
System.out.println(finalValue);
}
public void functionWithPara(int num) {
System.out.println("The parameter " + num
+ " has been passed by the method");
}
但是为什么这里用final却又可以访问呢?
因为Java采用了一种copy local variable的方式来实现,也就是说把定义为final的局部变量拷贝过来用,而引用的也可以拿过来用,只是不能重新赋值。从而造成了可以access local variable的假象,而这个时候由于不能重新赋值,所以一般不会造成不可预料的事情发生