在查看var
功能后,查看了here:
我在使用JDK 10设置Eclipse / IntelliJ IDEA IDE时遇到了困难,因此请求具有可用Java 10环境的Stack Overflow用户提供帮助。
请考虑以下事项:
public class A {
public void someMethod() { ... }
}
public class B extends A{
@Override
public void someMethod() { ... }
}
...
...
...
var myA = new A(); // Works as expected
myA = new B(); // Expected to fail in compilation due to var being
// syntactic sugar for declaring an A type
myA = (A) (new B()); // Should work
myA.someMethod(); // The question - which someMethod implementation is called?
使用var
时,我希望JVM能够识别变量所包含的派生类类型。并且在执行myA.someMethod()时执行B:someMethod()而不是A:someMethod()。
确实如此吗?
答案 0 :(得分:9)
感谢nullpointer向在线Java 10编译器提供link,我得到了以下有趣的结果:
public class Main {
static class A {
public void someMethod() { System.out.println(this.getClass().getName()); }
}
static class B extends A{
@Override
public void someMethod() { System.out.println("Derived: " + this.getClass().getName()); }
}
public static void main(String[] args) {
var myA = new A();
myA.someMethod();
myA = new B(); // does not fail to compile!
myA.someMethod();
}
}
输出:
Main$A // As expected
Derived: Main$B // As expected in inheritance
结论 - var是句法糖:
var myA = new A()
等同于A myA = new A()
,其中所有OOP都与之关联。
PS:我尝试用一个包含匿名类的var玩一点,并想出了这个有趣的行为 - 再次感谢nullpointer提到它是why can't we assign two inferred variables as an anonymous class to each other的副本:
static interface Inter {
public void method();
}
public static void main(String[] args) {
var inter = new Inter() {
@Override
public void method() {System.out.println("popo");}
};
inter.method();
inter = new Inter() {
@Override
public void method() {System.out.println("koko");}
};
inter.method();
}
输出:
Main.java:11: error: incompatible types: <anonymous Inter> cannot be converted to <anonymous Inter>
inter = new Inter() {
^
由于第二个匿名类类型与第一个匿名类类型不同,对var的第二个赋值失败 - 强制执行var
关键字的语法糖角色。
令人惊讶的是,错误消息不再精确 - 目前没有意义,因为错误中显示的类型名称是相同的!