DURGA SOFTWARE SOLUTIONS

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14. Internationalization

Question: 12
Given:
14. DateFormat df;
15. Date date = new Date();
16. //insert code here
17. String s = df.format( date);
Which two, inserted independently at line 16, allow the code to
compile? (Choose two.)
A. df= new DateFormat();
B. df= Date.getFormatter();
C. df= date.getFormatter();
D. df= date.getDateFormatter();
E. df= Date.getDateFormatter();
F. df= DateFormat.getInstance();
G. df = DateFormat.getDateInstance();
Answer: FG

Question: 13
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 will correctly split 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

14. Given:
import java.util.regex.*;
class Regex2 {
public static void main(String[] args) {
Pattern p = Pattern.compile(args[0]);
Matcher m = p.matcher(args[1]);
boolean b = false;
while(b = m.find()) {
System.out.print(m.start() + m.group());
}
}
}
And the command line:
java Regex2 "\d*" ab34ef
What is the result?
A. 234                                           B. 334
C. 2334                                          D. 0123456
E. 01234456                                  F. 12334567
G. Compilation fails.
Answer:
-> E is correct. The \d is looking for digits. The * is a quantifier that looks for 0 to many occurrences of the pattern that precedes it. Because we specified *, the group() method returns empty Strings until consecutive digits are found, so the only time group() returns a value is when it returns 34 when the matcher finds digits starting in position 2. The start() method returns the starting position of the previous match because, again, we said find 0 to many occurrences.
->A, B, C, D, E, F, and G are incorrect based on the above.

15. Given:
1. import java.text.*;
2. class DateOne {
3. public static void main(String[] args) {
4. Date d = new Date(1123631685981L);
5. DateFormat df = new DateFormat();
6. System.out.println(df.format(d));
7. }
8. }
And given that 1123631685981L is the number of milliseconds between Jan. 1, 1970, and
sometime on Aug. 9, 2005, what is the result? (Note: the time of day in option A may vary.)
A. 8/9/05 5:54 PM
B. 1123631685981L
C. An exception is thrown at runtime.
D. Compilation fails due to a single error in the code.
E. Compilation fails due to multiple errors in the code.
Answer:
-> E is correct. The Date class is located in the java.util package so it needs an import, and DateFormat objects must be created using a static method such as
DateFormat.getInstance() or DateFormat.getDateInstance().
->A, B, C, and D are incorrect based on the above.

16. Which are true? (Choose all that apply.)
A. The DateFormat.getDate() is used to convert a String to a Date instance.
B. Both DateFormat and NumberFormat objects can be constructed to be Locale specific.
C. Both Currency and NumberFormat objects must be constructed using static methods.
D. If a NumberFormat instance's Locale is to be different than the current Locale, it must be
specified at creation time.
E. A single instance of NumberFormat can be used to create Number objects from Strings and
to create formatted numbers from numbers.
Answer:
-> B , C, D, and E are correct.
->A is incorrect, DateFormat.parse() is used to convert a String to a Date.

17. Which will compile and run without exception? (Choose all that apply.)
A. System.out.format("%b", 123);
B. System.out.format("%c", "x");
C. System.out.printf("%d", 123);
D. System.out.printf("%f", 123);
E. System.out.printf("%d", 123.45);
F. System.out.printf("%f", 123.45);
G. System.out.format("%s", new Long("123"));
Answer:
->A , C, F, and G are correct. The %b (boolean) conversion character returns true for any non-null or non-boolean argument.
->B is incorrect, the %c (character) conversion character expects a character, not a String. D is incorrect, the %f (floating-point) conversion character won't automatically promote an integer type. E is incorrect, the %d (integral) conversion character won't take a floatingpoint number. (Note: The format() and printf() methods behave identically.)

18. Given:
1. import java.util.*;
2. class Brain {
3. public static void main(String[] args) {
4. // insert code block here
5. }
6. }
Which, inserted independently at line 4, compile and produce the output "123 82"?
(Choose all that apply.)
A. Scanner sc = new Scanner("123 A 3b c,45, x5x,76 82 L");
while(sc.hasNextInt()) System.out.print(sc.nextInt() + " ");
B. Scanner sc = new Scanner("123 A 3b c,45, x5x,76 82 L").
DURGA SCJP

useDelimiter(" ");
while(sc.hasNextInt()) System.out.print(sc.nextInt() + " ");
C. Scanner sc = new Scanner("123 A 3b c,45, x5x,76 82 L");
while(sc.hasNext()) {
if(sc.hasNextInt()) System.out.print(sc.nextInt() + " ");
else sc.next(); }
D. Scanner sc = new Scanner("123 A 3b c,45, x5x,76 82 L").
useDelimiter(" ");
while(sc.hasNext()) {
if(sc.hasNextInt()) System.out.print(sc.nextInt() + " ");
else sc.next(); }
E. Scanner sc = new Scanner("123 A 3b c,45, x5x,76 82 L");
do {
if(sc.hasNextInt()) System.out.print(sc.nextInt() + " ");
} while ( sc.hasNext() );
F. Scanner sc = new Scanner("123 A 3b c,45, x5x,76 82 L").
useDelimiter(" ");
do {
if(sc.hasNextInt()) System.out.print(sc.nextInt() + " ");
} while ( sc.hasNext() );
Answer:
->C and D are correct. Whitespace is the default delimiter, and the while loop advances through the String using nextInt() or next().
->A and B are incorrect because the while loop won't progress past the first non-int.
E and F are incorrect. The do loop will loop endlessly once the first non-int is found
because hasNext() does not advance through data.

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