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Oracle 1Z0-805 Exam Sample Questions

Question: 5

Given the code fragment:

public static void main(String[] args) {

String source = “d:\company\info.txt”;

String dest = “d:\company\emp\info.txt”;

//insert code fragment here Line **

} catch (IOException e) {

System.err.println (“Caught IOException: ” + e.getmessage();

}

}

Which two try statements, when inserted at line **, enable the code to successfully move the file info.txt to the destination directory, even if a file by the same name already exists in the destination directory?

  1. try {FileChannel in = new FileInputStream(source).getChannel();
  2. FileChannel out = new FileOutputStream(dest).getChannel ();

    in.transferTo (0, in.size(), out);

  3. try {Files.copy(Paths.get(source), Paths.get(dest));
  4. Files.delete(Paths.get(source));

  5. try {Files.copy(Paths.get(source), Paths.get(dest));
  6. Files.delete(Paths.get(source));

  7. try {Files.move (Paths.get(source), Paths.get(dest));
  8. try {BufferedReader br = Files.newBufferedReader(Paths.get(source), Charset.forName (“UTF-8”));

BufferedWriter bw = Files.newBufferedWriter (Paths.get(dest), Charset.forName (“UTF-8”));

String record = “”;

while ((record = br.readLine()) != null){

bw.write (record);

bw.newLine();

}

Files.delete(Paths.get(source));


Answer: B, D

Explanation:

Incorrect answers:

A: Copies (not moving) the file, but the original file is left.

C: Moves a file, but only if the destination does not exist.

  • The file is moved fine, but the content of the file is lost.
  • Question: 4

    Which code fragments print 1?

    1. String arr [] = {“1”, “2”, “3”};
    2. List arrList = new LinkedList <> (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    3. String arr [] = {“1”, “2”, “3”};
    4. List arrList = new LinkedList <> (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    5. String arr [] = {“1”, “2”, “3”};
    6. List arrList = new LinkedList <> (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    7. String arr [] = {“1”, “2”, “3”};
    8. List arrList = new LinkedList (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    9. String arr [] = {“1”, “2”, “3”};

    List extends String > arrList = new LinkedList (Arrays.asList (arr));

    System.out.println (arrList.get (0));


    Answer: A, C

    Explanation:

    Note: You can replace the type arguments required to invoke the constructor of a generic class with an empty set of type parameters (<>) as long as the compiler can infer the type arguments from the context. This pair of angle brackets is informally called the diamond.

    Incorrect answers:

    B: The Array is of type char. The List is of type Integer. Incompatible types.

    E: Type mismatch (Integer and char).

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    Question: 4

    Which code fragments print 1?

    1. String arr [] = {“1”, “2”, “3”};
    2. List arrList = new LinkedList <> (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    3. String arr [] = {“1”, “2”, “3”};
    4. List arrList = new LinkedList <> (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    5. String arr [] = {“1”, “2”, “3”};
    6. List arrList = new LinkedList <> (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    7. String arr [] = {“1”, “2”, “3”};
    8. List arrList = new LinkedList (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    9. String arr [] = {“1”, “2”, “3”};

    List extends String > arrList = new LinkedList (Arrays.asList (arr));

    System.out.println (arrList.get (0));


    Answer: A, C

    Explanation:

    Note: You can replace the type arguments required to invoke the constructor of a generic class with an empty set of type parameters (<>) as long as the compiler can infer the type arguments from the context. This pair of angle brackets is informally called the diamond.

    Incorrect answers:

    B: The Array is of type char. The List is of type Integer. Incompatible types.

    E: Type mismatch (Integer and char).

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    Question: 3

    Which two statements are true about the walkFileTree method of the files class?

    1. The file tree traversal is breadth-first with the given FileVisitor invoked for each file encountered.
    2. If the file is a directory, and if that directory could not be opened, the postVisitFileFailed method is invoked with the I/O exception.
    3. The maxDepth parameter’s value is the maximum number of directories to visit.
    4. By default, symbolic links are not automatically followed by the method.


    Answer: C, D

    Explanation:

    C: The method walkFileTree(Path start, Set options, int maxDepth, FileVisitor visitor)

    walks a file tree.

    The maxDepth parameter is the maximum number of levels of directories to visit. A value of 0 means that only the starting file is visited, unless denied by the security manager. A value of MAX_VALUE may be used to indicate that all levels should be visited. The visitFile method is invoked for all files, including directories, encountered at maxDepth, unless the basic file attributes cannot be read, in which case the visitFileFailed method is invoked.

    D: You need to decide whether you want symbolic links to be followed. If you are deleting files, for example, following symbolic links might not be advisable. If you are copying a file tree, you might want to allow it. By default, walkFileTree does not follow symbolic links.

    Incorrect answers:

    A: A file tree is walked depth first, but you cannot make any assumptions about the iteration order that subdirectories are visited.

    B: The method visitFileFailed(T file, IOException exc) is invoked for a file that could not be visited. This method is invoked if the file’s attributes could not be read, the file is a directory that could not be opened, and other reasons. However, there is no method named postVisitFileFailed.

    Reference: The Java Tutorials, Walking the File Tree

    Reference: walkFileTree

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    Question: 5

    Given the code fragment:

    public static void main(String[] args) {

    String source = “d:\company\info.txt”;

    String dest = “d:\company\emp\info.txt”;

    //insert code fragment here Line **

    } catch (IOException e) {

    System.err.println (“Caught IOException: ” + e.getmessage();

    }

    }

    Which two try statements, when inserted at line **, enable the code to successfully move the file info.txt to the destination directory, even if a file by the same name already exists in the destination directory?

    1. try {FileChannel in = new FileInputStream(source).getChannel();
    2. FileChannel out = new FileOutputStream(dest).getChannel ();

      in.transferTo (0, in.size(), out);

    3. try {Files.copy(Paths.get(source), Paths.get(dest));
    4. Files.delete(Paths.get(source));

    5. try {Files.copy(Paths.get(source), Paths.get(dest));
    6. Files.delete(Paths.get(source));

    7. try {Files.move (Paths.get(source), Paths.get(dest));
    8. try {BufferedReader br = Files.newBufferedReader(Paths.get(source), Charset.forName (“UTF-8”));

    BufferedWriter bw = Files.newBufferedWriter (Paths.get(dest), Charset.forName (“UTF-8”));

    String record = “”;

    while ((record = br.readLine()) != null){

    bw.write (record);

    bw.newLine();

    }

    Files.delete(Paths.get(source));


    Answer: B, D

    Explanation:

    Incorrect answers:

    A: Copies (not moving) the file, but the original file is left.

    C: Moves a file, but only if the destination does not exist.

  • The file is moved fine, but the content of the file is lost.
  • Question: 4

    Which code fragments print 1?

    1. String arr [] = {“1”, “2”, “3”};
    2. List arrList = new LinkedList <> (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    3. String arr [] = {“1”, “2”, “3”};
    4. List arrList = new LinkedList <> (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    5. String arr [] = {“1”, “2”, “3”};
    6. List arrList = new LinkedList <> (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    7. String arr [] = {“1”, “2”, “3”};
    8. List arrList = new LinkedList (Arrays.asList (arr));

      System.out.println (arrList.get (0));

    9. String arr [] = {“1”, “2”, “3”};

    List extends String > arrList = new LinkedList (Arrays.asList (arr));

    System.out.println (arrList.get (0));


    Answer: A, C

    Explanation:

    Note: You can replace the type arguments required to invoke the constructor of a generic class with an empty set of type parameters (<>) as long as the compiler can infer the type arguments from the context. This pair of angle brackets is informally called the diamond.

    Incorrect answers:

    B: The Array is of type char. The List is of type Integer. Incompatible types.

    E: Type mismatch (Integer and char).