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）