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- Question
Which will legally declare, construct, and initialize an array?
Options- A. int [] myList = {"1", "2", "3"};
- B. int [] myList = (5, 8, 2);
- C. int myList [] [] = {4,9,7,0};
- D. int myList [] = {4, 3, 7};
- Correct Answer
- int myList [] = {4, 3, 7};
ExplanationThe only legal array declaration and assignment statement is Option DOption A is wrong because it initializes an int array with String literals.
Option B is wrong because it use something other than curly braces for the initialization.
Option C is wrong because it provides initial values for only one dimension, although the declared array is a two-dimensional array.
Language Fundamentals problems
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- 1. 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).
- 2. 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.
- 3. 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.
- 4. Which three are valid declarations of a char?
- char c1 = 064770;
- char c2 = 'face';
- char c3 = 0xbeef;
- char c4 = \u0022;
- char c5 = '\iface';
- char c6 = '\uface';
Options- A. 1, 2, 4
- B. 1, 3, 6
- C. 3, 5
- D. 5 only Discuss
Correct Answer: 1, 3, 6
Explanation:
(1), (3), and (6) are correct. char c1 = 064770; is an octal representation of the integer value 27128, which is legal because it fits into an unsigned 16-bit integer. char c3 = 0xbeef; is a hexadecimal representation of the integer value 48879, which fits into an unsigned 16-bit integer. char c6 = '\uface'; is a Unicode representation of a character.char c2 = 'face'; is wrong because you can't put more than one character in a char literal. The only other acceptable char literal that can go between single quotes is a Unicode value, and Unicode literals must always start with a '\u'.
char c4 = \u0022; is wrong because the single quotes are missing.
char c5 = '\iface'; is wrong because it appears to be a Unicode representation (notice the backslash), but starts with '\i' rather than '\u'.
- 5. Which is a reserved word in the Java programming language?
Options- A. method
- B. native
- C. subclasses
- D. reference
- E. array Discuss
Correct Answer: native
Explanation:
The word "native" is a valid keyword, used to modify a method declaration.Option A, D and E are not keywords. Option C is wrong because the keyword for subclassing in Java is extends, not 'subclasses'.
- 6. 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.- 7. What will be the output of the program?
class Bitwise { public static void main(String [] args) { int x = 11 & 9; int y = x ^ 3; System.out.println( y | 12 ); } }
Options- A. 0
- B. 7
- C. 8
- D. 14 Discuss
Correct Answer: 14
Explanation:
The & operator produces a 1 bit when both bits are 1. The result of the & operation is 9. The ^ operator produces a 1 bit when exactly one bit is 1; the result of this operation is 10. The | operator produces a 1 bit when at least one bit is 1; the result of this operation is 14.- 8. What will be the output of the program?
class Two { byte x; } class PassO { public static void main(String [] args) { PassO p = new PassO(); p.start(); } void start() { Two t = new Two(); System.out.print(t.x + " "); Two t2 = fix(t); System.out.println(t.x + " " + t2.x); } Two fix(Two tt) { tt.x = 42; return tt; } }
Options- A. null null 42
- B. 0 0 42
- C. 0 42 42
- D. 0 0 0 Discuss
Correct Answer: 0 42 42
Explanation:
In the fix() method, the reference variable tt refers to the same object (class Two) as the t reference variable. Updating tt.x in the fix() method updates t.x (they are one in the same object). Remember also that the instance variable x in the Two class is initialized to 0.- 9. What will be the output of the program?
class Equals { public static void main(String [] args) { int x = 100; double y = 100.1; boolean b = (x = y); /* Line 7 */ System.out.println(b); } }
Options- A. true
- B. false
- C. Compilation fails
- D. An exception is thrown at runtime Discuss
Correct Answer: Compilation fails
Explanation:
The code will not compile because in line 7, the line will work only if we use (x==y) in the line. The == operator compares values to produce a boolean, whereas the = operator assigns a value to variables.Option A, B, and D are incorrect because the code does not get as far as compiling. If we corrected this code, the output would be false.
- 10. What will be the output of the program?
class Test { public static void main(String [] args) { int x=20; String sup = (x < 15)? "small" : (x < 22)? "tiny" : "huge"; System.out.println(sup); } }
Options- A. small
- B. tiny
- C. huge
- D. Compilation fails Discuss
Correct Answer: tiny
Explanation:
This is an example of a nested ternary operator. The second evaluation (x < 22) is true, so the "tiny" value is assigned to sup.
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- 1. Which three are legal array declarations?