Java基础、面试知识点

All objects are allocated on heap in Java

问题1：(继承)

class Base {    public static void show() {       System.out.println("Base::show() called");    }}  class Derived extends Base {    public static void show() {       System.out.println("Derived::show() called");    }}  class Main {    public static void main(String[] args) {        Base b = new Derived();;        b.show();    }}

输出结果： (A) Base::show() called (B)Derived::show() calle (C) 编译错误

问题2：(继承)

class Base {    public void foo() { System.out.println("Base"); }}  class Derived extends Base {    private void foo() { System.out.println("Derived"); }}  public class Main {    public static void main(String args[]) {        Base b = new Derived();        b.foo();    }}

输出结果：(A) Base (B) Derived (C) 编译错误 (D)运行错误

问题3：(继承)

class Grandparent {    public void Print() {        System.out.println("Grandparent's Print()");    }}  class Parent extends Grandparent {    public void Print() {        System.out.println("Parent's Print()");    }}  class Child extends Parent {    public void Print() {        super.super.Print();         System.out.println("Child's Print()");    }}  public class Main {    public static void main(String[] args) {        Child c = new Child();        c.Print();    }}

输出结果：(A) 编译错误  (B)Grandparent's Print() Parent's Print() Child's Print() (C) 运行错误

问题4：(方法)

class Test {public static void swap(Integer i, Integer j) {      Integer temp = new Integer(i);      i = j;      j = temp;   }   public static void main(String[] args) {      Integer i = new Integer(10);      Integer j = new Integer(20);      swap(i, j);      System.out.println("i = " + i + ", j = " + j);   }}

输出结果：(A)i = 10, j = 20 (B)i = 20, j = 10 (C)i = 10, j = 10 (D)i = 20, j = 20

问题5：(异常)

class Base extends Exception {}class Derived extends Base  {} public class Main {  public static void main(String args[]) {   // some other stuff   try {       // Some monitored code       throw new Derived();    }    catch(Base b)     {        System.out.println("Caught base class exception");     }    catch(Derived d)  {        System.out.println("Caught derived class exception");     }  }}

输出结果：(A)Caught base class exception (B)Caught derived class exception (C)编译错误 because derived is not throwable (D)编译错误 because base class exception is caught before derived class

问题6：(操作符)

class Test {    public static void main(String args[])  {       int x = -4;       System.out.println(x>>1);        int y = 4;       System.out.println(y>>1);     }  }

输出结果是？

问题7：(操作符)

class Base {} class Derived extends Base {   public static void main(String args[]){      Base a = new Derived();      System.out.println(a instanceof Derived);   }}

输出结果是？

解析

问题1：选A。方法是static的，不会出现运行时多态Like C++, when a function is static, runtime polymorphism doesn't happen.
问题2：选C。子类覆盖父类的方法，不应比父类方法的限定修饰词更严格。It is compiler error to give more restrictive access to a derived class function which overrides a base class function.

问题3：选A。JAVA里 super.super是不允许的。In Java, it is not allowed to do super.super. We can only access Grandparent's members using Parent. 

问题4：选A。JAVA中参数通过值传递。Parameters are passed by value in Java

问题5：选D。

Main.java:12: error: exception Derived has already been caught    catch(Derived d)  { System.out.println("Caught derived class exception"); } 

问题6：-2 2 带符号右移（算术右移）。知乎有个答案不错：https://www.zhihu.com/question/38051371

问题7：true  当引用为父类类型时，instanceof操作符同样成立。

（未完待续）