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- Question
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
- Correct Answer
- Compilation fails
ExplanationThe 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.
Operators and Assignments problems
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- 1. 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.- 2. 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.- 3. 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.- 4. 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}; Discuss
Correct Answer: int myList [] = {4, 3, 7};
Explanation:
The 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.
- 5. 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).
- 6. 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.- 7. What will be the output of the program?
class Test { public static void main(String [] args) { Test p = new Test(); p.start(); } void start() { boolean b1 = false; boolean b2 = fix(b1); System.out.println(b1 + " " + b2); } boolean fix(boolean b1) { b1 = true; return b1; } }
Options- A. true true
- B. false true
- C. true false
- D. false false Discuss
Correct Answer: false true
Explanation:
The boolean b1 in the fix() method is a different boolean than the b1 in the start() method. The b1 in the start() method is not updated by the fix() method.- 8. What will be the output of the program?
class BitShift { public static void main(String [] args) { int x = 0x80000000; System.out.print(x + " and "); x = x >>> 31; System.out.println(x); } }
Options- A. -2147483648 and 1
- B. 0x80000000 and 0x00000001
- C. -2147483648 and -1
- D. 1 and -2147483648 Discuss
Correct Answer: -2147483648 and 1
Explanation:
Option A is correct. The >>> operator moves all bits to the right, zero filling the left bits. The bit transformation looks like this:Before: 1000 0000 0000 0000 0000 0000 0000 0000
After: 0000 0000 0000 0000 0000 0000 0000 0001
Option C is incorrect because the >>> operator zero fills the left bits, which in this case changes the sign of x, as shown.
Option B is incorrect because the output method print() always displays integers in base 10.
Option D is incorrect because this is the reverse order of the two output numbers.
- 9. What will be the output of the program?
class Test { static int s; public static void main(String [] args) { Test p = new Test(); p.start(); System.out.println(s); } void start() { int x = 7; twice(x); System.out.print(x + " "); } void twice(int x) { x = x*2; s = x; } }
Options- A. 7 7
- B. 7 14
- C. 14 0
- D. 14 14 Discuss
Correct Answer: 7 14
Explanation:
The int x in the twice() method is not the same int x as in the start() method. Start()'s x is not affected by the twice() method. The instance variable s is updated by twice()'s x, which is 14.- 10. What will be the output of the program?
class SSBool { public static void main(String [] args) { boolean b1 = true; boolean b2 = false; boolean b3 = true; if ( b1 & b2 | b2 & b3 | b2 ) /* Line 8 */ System.out.print("ok "); if ( b1 & b2 | b2 & b3 | b2 | b1 ) /*Line 10*/ System.out.println("dokey"); } }
Options- A. ok
- B. dokey
- C. ok dokey
- D. No output is produced
- E. Compilation error Discuss
Correct Answer: dokey
Explanation:
The & operator has a higher precedence than the | operator so that on line 8 b1 and b2 are evaluated together as are b2 & b3. The final b1 in line 10 is what causes that if test to be true. Hence it prints "dokey".
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- 1. What will be the output of the program?