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- Question
You need to store elements in a collection that guarantees that no duplicates are stored and all elements can be accessed in natural order. Which interface provides that capability?
Options- A. java.util.Map
- B. java.util.Set
- C. java.util.List
- D. java.util.Collection
- Correct Answer
- java.util.Set
ExplanationOption B is correct. A set is a collection that contains no duplicate elements. The iterator returns the elements in no particular order (unless this set is an instance of some class that provides a guarantee). A map cannot contain duplicate keys but it may contain duplicate values. List and Collection allow duplicate elements.Option A is wrong. A map is an object that maps keys to values. A map cannot contain duplicate keys; each key can map to at most one value. The Map interface provides three collection views, which allow a map's contents to be viewed as a set of keys, collection of values, or set of key-value mappings. The order of a map is defined as the order in which the iterators on the map's collection views return their elements. Some map implementations, like the TreeMap class, make specific guarantees as to their order (ascending key order); others, like the HashMap class, do not (does not guarantee that the order will remain constant over time).
Option C is wrong. A list is an ordered collection (also known as a sequence). The user of this interface has precise control over where in the list each element is inserted. The user can access elements by their integer index (position in the list), and search for elements in the list. Unlike sets, lists typically allow duplicate elements.
Option D is wrong. A collection is also known as a sequence. The user of this interface has precise control over where in the list each element is inserted. The user can access elements by their integer index (position in the list), and search for elements in the list. Unlike sets, lists typically allow duplicate elements.
More questions
- 1. Which of the following statements is true?
Options- A. In an assert statement, the expression after the colon ( : ) can be any Java expression.
- B. If a switch block has no default, adding an assert default is considered appropriate.
- C. In an assert statement, if the expression after the colon ( : ) does not have a value, the assert's error message will be empty.
- D. It is appropriate to handle assertion failures using a catch clause. Discuss
Correct Answer: If a switch block has no default, adding an assert default is considered appropriate.
Explanation:
Adding an assertion statement to a switch statement that previously had no default case is considered an excellent use of the assert mechanism.Option A is incorrect because only Java expressions that return a value can be used. For instance, a method that returns void is illegal.
Option C is incorrect because the expression after the colon must have a value.
Option D is incorrect because assertions throw errors and not exceptions, and assertion errors do cause program termination and should not be handled.
- 2. Given a method in a protected class, what access modifier do you use to restrict access to that method to only the other members of the same class?
Options- A. final
- B. static
- C. private
- D. protected
- E. volatile Discuss
Correct Answer: private
Explanation:
The private access modifier limits access to members of the same class.Option A, B, D, and E are wrong because protected are the wrong access modifiers, and final, static, and volatile are modifiers but not access modifiers.
- 3. Which is a valid declaration within an interface?
Options- A. public static short stop = 23;
- B. protected short stop = 23;
- C. transient short stop = 23;
- D. final void madness(short stop); Discuss
Correct Answer: public static short stop = 23;
Explanation:
(A) is valid interface declarations.(B) and (C) are incorrect because interface variables cannot be either protected or transient. (D) is incorrect because interface methods cannot be final or static.
- 4. Which four options describe the correct default values for array elements of the types indicated?
- int -> 0
- String -> "null"
- Dog -> null
- char -> '\u0000'
- float -> 0.0f
- boolean -> true
Options- A. 1, 2, 3, 4
- B. 1, 3, 4, 5
- C. 2, 4, 5, 6
- D. 3, 4, 5, 6 Discuss
Correct Answer: 1, 3, 4, 5
Explanation:
(1), (3), (4), (5) are the correct statements.(2) is wrong because the default value for a String (and any other object reference) is null, with no quotes.
(6) is wrong because the default value for boolean elements is false.
- 5. Which one of these lists contains only Java programming language keywords?
Options- A. class, if, void, long, Int, continue
- B. goto, instanceof, native, finally, default, throws
- C. try, virtual, throw, final, volatile, transient
- D. strictfp, constant, super, implements, do
- E. byte, break, assert, switch, include Discuss
Correct Answer: goto, instanceof, native, finally, default, throws
Explanation:
All the words in option B are among the 49 Java keywords. Although goto reserved as a keyword in Java, goto is not used and has no function.Option A is wrong because the keyword for the primitive int starts with a lowercase i.
Option C is wrong because "virtual" is a keyword in C++, but not Java.
Option D is wrong because "constant" is not a keyword. Constants in Java are marked static and final.
Option E is wrong because "include" is a keyword in C, but not in Java.
- 6. Which three are legal array declarations?
- int [] myScores [];
- char [] myChars;
- int [6] myScores;
- Dog myDogs [];
- Dog myDogs [7];
Options- A. 1, 2, 4
- B. 2, 4, 5
- C. 2, 3, 4
- D. All are correct. Discuss
Correct Answer: 1, 2, 4
Explanation:
(1), (2), and (4) are legal array declarations. With an array declaration, you can place the brackets to the right or left of the identifier. Option A looks strange, but it's perfectly legal to split the brackets in a multidimensional array, and place them on both sides of the identifier. Although coding this way would only annoy your fellow programmers, for the exam, you need to know it's legal.(3) and (5) are wrong because you can't declare an array with a size. The size is only needed when the array is actually instantiated (and the JVM needs to know how much space to allocate for the array, based on the type of array and the size).
- 7. What will be the output of the program?
class SC2 { public static void main(String [] args) { SC2 s = new SC2(); s.start(); } void start() { int a = 3; int b = 4; System.out.print(" " + 7 + 2 + " "); System.out.print(a + b); System.out.print(" " + a + b + " "); System.out.print(foo() + a + b + " "); System.out.println(a + b + foo()); } String foo() { return "foo"; } }
Options- A. 9 7 7 foo 7 7foo
- B. 72 34 34 foo34 34foo
- C. 9 7 7 foo34 34foo
- D. 72 7 34 foo34 7foo Discuss
Correct Answer: 72 7 34 foo34 7foo
Explanation:
Because all of these expressions use the + operator, there is no precedence to worry about and all of the expressions will be evaluated from left to right. If either operand being evaluated is a String, the + operator will concatenate the two operands; if both operands are numeric, the + operator will add the two operands.- 8. What will be the output of the program?
class Test { public static void main(String [] args) { int x= 0; int y= 0; for (int z = 0; z < 5; z++) { if (( ++x > 2 ) || (++y > 2)) { x++; } } System.out.println(x + " " + y); } }
Options- A. 5 3
- B. 8 2
- C. 8 3
- D. 8 5 Discuss
Correct Answer: 8 2
Explanation:
The first two iterations of the for loop both x and y are incremented. On the third iteration x is incremented, and for the first time becomes greater than 2. The short circuit or operator || keeps y from ever being incremented again and x is incremented twice on each of the last three iterations.- 9. What will be the output of the program?
public class Test { public static void leftshift(int i, int j) { i <<= j; } public static void main(String args[]) { int i = 4, j = 2; leftshift(i, j); System.out.printIn(i); } }
Options- A. 2
- B. 4
- C. 8
- D. 16 Discuss
Correct Answer: 4
Explanation:
Java only ever passes arguments to a method by value (i.e. a copy of the variable) and never by reference. Therefore the value of the variable i remains unchanged in the main method.If you are clever you will spot that 16 is 4 multiplied by 2 twice, (4 * 2 * 2) = 16. If you had 16 left shifted by three bits then 16 * 2 * 2 * 2 = 128. If you had 128 right shifted by 2 bits then 128 / 2 / 2 = 32. Keeping these points in mind, you don't have to go converting to binary to do the left and right bit shifts.
- 10. What will be the output of the program?
class PassS { public static void main(String [] args) { PassS p = new PassS(); p.start(); } void start() { String s1 = "slip"; String s2 = fix(s1); System.out.println(s1 + " " + s2); } String fix(String s1) { s1 = s1 + "stream"; System.out.print(s1 + " "); return "stream"; } }
Options- A. slip stream
- B. slipstream stream
- C. stream slip stream
- D. slipstream slip stream Discuss
Correct Answer: slipstream slip stream
Explanation:
When the fix() method is first entered, start()'s s1 and fix()'s s1 reference variables both refer to the same String object (with a value of "slip"). Fix()'s s1 is reassigned to a new object that is created when the concatenation occurs (this second String object has a value of "slipstream"). When the program returns to start(), another String object is created, referred to by s2 and with a value of "stream".
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