OCJP标准题库试题之————第二次

深藏阁楼爱情的钟 2022-08-18 08:17 150阅读 0赞

QUESTION 13
    Click the Exhibit button. Which statement is true about the classes and interfaces in the exhibit?

![这里写图片描述][20151210155213384]

A. Compilation will succeed for all classes and interfaces.
    B. Compilation of class C will fail because of an error in line 2.
    C. Compilation of class C will fail because of an error in line 6.
    D. Compilation of class AImpl will fail because of an error in line 2.
    Answer: C
    Section: (none)

**我们要明白一个知识点：继承重写的返回值。可以是父类但不能是更高一级别的父类。子类的抛的异常不能是父类的异常。可以是向下抛出子类的异常。**

**我们看到了，AImpl实现了A接口。C类继承 B类，但是C类重写B类的execute方法的时候，有问题，C类重写方法的返回值是B类中execute的返回值的父类。这是错误的。编译不能通过的。由此从选项中，分析：A：所有的类和接口都编译成功 显然是不对的。**

**B：C类编译失败。错误在第二行。根据我们分析错误实在第6行。B×**

**C：编译失败。错误在第六行。C√**

**D：编辑AImpl会失败，因为错误在第二行。D×**

QUESTION 14
    Which two code fragments correctly create and initialize a static array of int elements? (Choose two.)
    A. static final int[] a = { 100,200 };
    B. static final int[] a;
    static { a=new int[2]; a[0]=100; a[1]=200; }
    C. static final int[] a = new int[2]{ 100,200 };
    D. static final int[] a;
    static void init() { a = new int[3]; a[0]=100; a[1]=200; }
    Answer: AB
    Section: (none)

**分析：这里考察的是对象的初始化,常见的初始化的方法：1.构造方法，2.初始化块，3静态块。他们的运行优先性是 3》2》1。**

**A：A项是生命即初始化，正确**  
**B:先声明，后初始化，采用的是静态块初始化,B√**  
**C：标注长度，画蛇添足。C×**  
**D：根本没有这样的写法**

QUESTION 15
    Given:
    10. interface Foo { int bar(); }
    11. public class Sprite {
    12. public int fubar( Foo foo ) { return foo.bar(); }
    13. public void testFoo() {
    14. fubar(
    15. // insert code here
    16. );
    17. }
    18. }
    Which code, inserted at line 15, allows the class Sprite to compile?
    A. Foo { public int bar() { return 1; }
    B. new Foo { public int bar() { return 1; }
    C. new Foo() { public int bar() { return 1; }
    D. new class Foo { public int bar() { return 1; }
    Answer: C

**分析：这里考察的是匿名类的使用。让我们在15行写代码让这个类能编译。显然选择C。只有C让通过编译。一个接口没有实现类，而另一个方法的参数又是这个接口，调用里面的方法。然后又调用了这个方法。这时候就需要new这个接口。并且在里面实现该接口的方法。**

QUESTION 16
    Given:
    1. class Alligator {
    2. public static void main(String[] args) {
    3. int []x[] = {
       {1,2}, {3,4,5}, {6,7,8,9}};
    4. int [][]y = x;
    5. System.out.println(y[2][1]);
    6. }
    7. }
    What is the result?
    A. 2
    B. 3
    C. 4
    D. 6
    E. 7
    F. Compilation fails.
    Answer: E

**这里有一个锯齿数组：内容 第一行 1，2 第二行 3，4，5.第三行，6，7，8，9. 求Y【2】【1】. 数字从0开始。所以我们要找的值是第三行第二列的7.选择E**

QUESTION 17
    Given:
    22. StringBuilder sb1 = new StringBuilder("123");
    23. String s1 = "123";
    24. // insert code here
    25. System.out.println(sb1 + " " + s1);
    Which code fragment, inserted at line 24, outputs "123abc 123abc"?
    A. sb1.append("abc"); s1.append("abc");
    B. sb1.append("abc"); s1.concat("abc");
    C. sb1.concat("abc"); s1.append("abc");
    D. sb1.concat("abc"); s1.concat("abc");
    E. sb1.append("abc"); s1 = s1.concat("abc");
    F. sb1.concat("abc"); s1 = s1.concat("abc");
    G. sb1.append("abc"); s1 = s1 + s1.concat("abc");
    H. sb1.concat("abc"); s1 = s1 + s1.concat("abc");
    Answer: E

**分析：考察StringBuffer的append方法和String的contact方法：append追加字符串。contact方法：一个字符串拼接另一个字符串，返回拼接的结果,不修改原来的值。根据这个知识我们分析每个选项的sb1+s1的结果**

**A：String类型没有append方法**  
**B：123abc+123 B×**  
**C：同A**  
**D：Stringbuffer没有concat方法**  
**E:正确**  
**F：同D**  
**G：123abc+123123abc 也不对**  
**H:同D**

QUESTION 18
    Given that the current directory is empty, and that the user has read and write permissions, and the
    following:
    11. import java.io.*;
    12. public class DOS {
    13. public static void main(String[] args) {
    14. File dir = new File("dir");
    15. dir.mkdir();
    16. File f1 = new File(dir, "f1.txt");
    17. try {
    18. f1.createNewFile();
    19. } catch (IOException e) { ; }
    20. File newDir = new File("newDir");
    21. dir.renameTo(newDir);
    22. }
    23. }
    Which statement is true?
    A. Compilation fails.
    B. The file system has a new empty directory named dir.
    C. The file system has a new empty directory named newDir.
    D. The file system has a directory named dir, containing a file f1.txt.
    E. The file system has a directory named newDir, containing a file f1.txt.
    Answer: E

**分析题目：编译正确：15行：创建了一个叫做dir的目录，18行在dir目录创建了一个f1.txt的文件。21行将dir目录改名为newDir。所以钻则E。文件系统有个叫做newdir的目录里面有个f1.txt的文件**

QUESTION 19
    Given:
    11. class Converter {
    12. public static void main(String[] args) {
    13. Integer i = args[0];
    14. int j = 12;
    15. System.out.println("It is " + (j==i) + " that j==i.");
    16. }
    17. }
    What is the result when the programmer attempts to compile the code and run it with the command java
    Converter 12?
    A. It is true that j==i.
    B. It is false that j==i.
    C. An exception is thrown at runtime.
    D. Compilation fails because of an error in line 13.
    Answer: D
    Section: (none)

**分析。main函数开始接受了字符串数组。12.字符串不会自动转为integer的。所以编译错误。**

QUESTION 20
    Given:
    11. String test = "Test A. Test B. Test C.";
    12. // insert code here
    13. String[] result = test.split(regex);
    Which regular expression, inserted at line 12, correctly splits test into "Test A", "Test B", and "Test C"?
    A. String regex = "";
    B. String regex = " ";
    C. String regex = ".*";
    D. String regex = "\\s";
    E. String regex = "\\.\\s*";
    F. String regex = "\\w[ \.] +";
    Answer: E
    Section: (none)

**考察 java的正则表达式。显然分析是用.和空格来分割这个字符串的。.指任意字符。s标识space空格。当然需要转义。\\有特殊的用途所以就双斜线。**

QUESTION 21
    Given:
    5. import java.util.Date;
    6. import java.text.DateFormat;
    21. DateFormat df;
    22. Date date = new Date();
    23. // insert code here
    24. String s = df.format(date);
    Which code fragment, inserted at line 23, allows the code to compile?
    A. df = new DateFormat();
    B. df = Date.getFormat();
    C. df = date.getFormat();
    D. df = DateFormat.getFormat();
    E. df = DateFormat.getInstance();
    Answer: E

**格式化：日期我们要把 格式化日期工具创建出来。DateFormat创建不是通过new实现的。选择E**

Given a class Repetition:
    1. package utils;
    2.
    3. public class Repetition { 
    4. public static String twice(String s) { return s + s; }
    5. } and given another class Demo: 1. // insert code here
    2.
    3. public class Demo { 
    4. public static void main(String[] args) {
    5. System.out.println(twice("pizza"));
    6. }
    7. }
    Which code should be inserted at line 1 of Demo.java to compile and run Demo to print "pizzapizza"?
    A. import utils.*;
    B. static import utils.*;
    C. import utils.Repetition.*;
    D. static import utils.Repetition.*;
    E. import utils.Repetition.twice();
    F. import static utils.Repetition.twice;
    G. static import utils.Repetition.twice;
    Answer: F

**这里考察的是静态导入。一个类的某个方法. import +static+包名+方法名字**

import static utils.Repetition.twice

QUESTION 23
    A UNIX user named Bob wants to replace his chess program with a new one, but he is not sure where the
    old one is installed. Bob is currently able to run a Java chess program starting from his home directory /
    home/bob using the command: java -classpath /test:/home/bob/downloads/*.jar games.Chess Bob's
    CLASSPATH is set (at login time) to:
    /usr/lib:/home/bob/classes:/opt/java/lib:/opt/java/lib/*.jar What is a possible location for the Chess.class file?
    A. /test/Chess.class
    B. /home/bob/Chess.class
    C. /test/games/Chess.class
    D. /usr/lib/games/Chess.class
    E. /home/bob/games/Chess.class
    F. inside jarfile /opt/java/lib/Games.jar (with a correct manifest)
    G. inside jarfile /home/bob/downloads/Games.jar (with a correct manifest)
    Answer: C
    Section: (none)

**这里考察的类路径。一个叫bob的人。有个象棋程序。现在要去更新原来的程序。但是不知道这个程序放哪了。现在能运行原来的程序。让我们猜测这个程序在哪？java -classpath /test:/home/bob/downloads/\*.jar games.Chess。通过这里。程序运行，加载test下的包。和/home/bob/downloads/下的所有jar文件。那么程序可能就在test下和downloads下。有因为。调用是 games.Chess，运行程序。downloads目录是一堆jar包。而调用在games目录下。所以可能就在/test/games/Chess.class这里。选择C**

其它题目参考 [http://www.cnblogs.com/husam/p/3880475.html][http_www.cnblogs.com_husam_p_3880475.html]

[20151210155213384]: /images/20220731/b9985926376a40c8b4e222df16d801da.png
[http_www.cnblogs.com_husam_p_3880475.html]: http://www.cnblogs.com/husam/p/3880475.html