Interview

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推断打印结果

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public static void main(String args[]) {
  String a = "hello2";
  final String b = "hello";
  String d = "hello";
  String c = b + 2;
  String e = d + 2;
  
  System.out.println((a == c));
  System.out.println((a == e));
}

结果：

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true
false

解析：由于b的值确定且被final修饰，会被当做编译期常量使用，即编译期出现b变量的地方会被直接替换，
替换后，a和c都指向字符串常量池中的hello2地址，故a==c为true；而String e = d + 2;会被编译器转换为类似于 String e = new StringBuilder(d).append(“2”); 故 a==e为false.

请画出a，b，c，d内存分布图

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public static void main(String args[]) {
  int[] a = { 1, 2, 3 };
  int[] b = { 4, 5, 6 };
  int[] c = { 7, 8, 9 };
  int[][] d = { a, b, c };
}

解析：

判断输出结果

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public class Example {
  String str = new String("Hello");
  char[] ch = {'a','b','c'};

  public static void main(String[] args) {
     Example ex = new Example();
     ex.change(ex.str, ex.ch);
     System.out.println(ex.str);
     System.out.println(ex.ch);
  }

  public void change(String str, char[] ch){
     str= "Hi";
     ch[1]= 'h';
  }
}

运行结果：

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Hello
ahc

解析：

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public class Person {

  public int age;

  public static void changeEmployee(Person employee) {
    employee = new Person();
    employee.age = 1000;
  }

  public static void main(String[] args) {
    Person employee = new Person();
    employee.age = 100;
    changeEmployee(employee);
    System.out.println(employee.age);
  }
}

哪些操作是原子性操作

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x = 10;         //语句1
y = x;          //语句2
x++;            //语句3
x = x + 1;      //语句4
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x = 10;

解析：注意y = x，需要读取和赋值，并非原子性操作。

以下代码是否有问题，并解释。

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public class Example {
  public static void main(String[] args) {
    Outputer outputer = new Outputer();
    final Thread t0 = new Thread(() -> {
      outputer.output2("How are you?");
    });
    final Thread t1 = new Thread(() -> {
      outputer.output("Fine, thank you.");
    });
    t0.start();
    t1.start();
  }

  static class Outputer {
    public synchronized void output(String name) {
      int len = name.length();
      for (int i = 0; i < len; i++) {
        System.out.print(name.charAt(i));
      }
      System.out.println();
    }

    public static synchronized void output2(String name) {
      int len = name.length();
      for (int i = 0; i < len; i++) {
        System.out.print(name.charAt(i));
      }
      System.out.println();
    }
  }
}

解析：
ouput和output2两个方法分别使用类对象和类实例对象的锁，两个线程不会竞争同一个锁资源，因而不会产生互斥，从而导致输出混乱。

SQL查询

员工表emp：

empno ename job mgr hiredate sal comm deptno
7369 SMITH CLERK 7902 1980-12-17 800.00 NULL 20
7499 ALLEN SALESMAN 7698 1981-02-20 1600.00 300.00 30
7521 WARD SALESMAN 7698 1981-02-22 1250.00 500.00 30
7566 JONES MANAGER 7839 1981-04-02 2975.00 NULL 20
7654 MARTIN SALESMAN 7698 1981-09-28 1250.00 1400.00 30
7698 BLAKE MANAGER 7839 1981-05-01 2850.00 NULL 30
7782 CLARK MANAGER 7839 1981-06-09 2450.00 NULL 10
7788 SCOTT ANALYST 7566 1987-07-13 3000.00 NULL 20
7839 KING PRESIDENT NULL 1981-11-17 5000.00 NULL 10
7844 TURNER SALESMAN 7698 1981-09-08 1500.00 0.00 30
7876 ADAMS CLERK 7788 1987-07-13 1100.00 NULL 20
7900 JAMES CLERK 7698 1981-12-03 950.00 NULL 30
7902 FORD ANALYST 7566 1981-12-03 3000.00 NULL 20
7934 MILLER CLERK 7782 1982-01-23 1300.00 NULL 10

部门表dept：

deptno dname loc
10 ACCOUNTING NEW YORK
20 RESEARCH DALLAS
30 SALES CHICAGO
40 OPERATIONS BOSTON
查询部门表在员工表中不存在的部门号;
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SELECT d.deptno
FROM dept d
WHERE NOT EXISTS (
SELECT NULL
FROM emp e
WHERE d.deptno = e.deptno);
查询工资高于部门平均工资的员工;
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SELECT e2.ename,e2.sal,e2.deptno,r1._avg
FROM emp e2
INNER JOIN (
SELECT e1.deptno, AVG(e1.sal) AS _avg
FROM emp e1
GROUP BY e1.deptno) r1 ON e2.deptno = r1.deptno AND e2.sal > r1._avg
ORDER BY e2.deptno ASC, e2.sal DESC;
查询部门工资前三名的员工;
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SELECT t0.deptno, t0.ename, t0.sal
FROM emp t0
WHERE (
SELECT COUNT(1)
FROM emp t1
WHERE t0.deptno = t1.deptno AND t1.sal >= t0.sal) <= 3
ORDER BY deptno,sal DESC;

解析： 员工工资排名前3，可以转换为: 工资大于等于自己的人数(count)小于等于3