visual basic barcode program 3: Operators and Assignments in Java

Printer PDF 417 in Java 3: Operators and Assignments

3: Operators and Assignments
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12. 13. 14. 15. 16. 17. 18.
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} long [] fix(long [] a3) { a3[1] = 7; return a3; } }
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A. 12 15 B. 15 15 C. 3 4 5 3 7 5 D. 3 7 5 3 7 5 E. Compilation fails F.
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18. Given the following,
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1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 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; } }
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Self Test
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A. null null 42 B. 0 0 42 C. 0 42 42 D. 0 0 0 E. Compilation fails F.
An exception is thrown at runtime
3: Operators and Assignments
SELF TEST ANSWERS
Java Operators (Sun Objective 5.1)
1. B and D. B and D both evaluate to 32. B is shifting bits right then left using the signed bit shifters >> and <<. D is shifting bits using the unsigned operator >>>, but since the beginning number is positive the sign is maintained. th A evaluates to 8, C looks like 2 to the 5 power, but ^ is the Exclusive OR operator so C th evaluates to 7. E evaluates to 16, and F evaluates to 0 (2 >> 5 is not 2 to the 5 ). 2. B and D. B is correct because class type Ticker is part of the class hierarchy of t; therefore it is a legal use of the instanceof operator. D is also correct because Component is part of the hierarchy of t, because Ticker extends Component in line 2. A is incorrect because the syntax is wrong. A variable (or null) always appears before the instanceof operator, and a type appears after it. C and E are incorrect because the statement is used as a method, which is illegal. F is incorrect because the String class is not in the hierarchy of the t object. 3. C. The code will not compile because in line 5, 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. 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. 4. B, D, and E. B is correct because the reference variables f1 and f3 refer to the same array object. D is correct because it is legal to compare integer and floating-point types. E is correct because it is legal to compare a variable with an array element. C is incorrect because f2 is an array object and f1[1] is an array element. 5. A. 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 C is incorrect because the >>> operator zero fills the left bits, which in this case changes the sign of x, as shown. B is incorrect because the output method print() always displays integers in base 10. D is incorrect because this is the reverse order of the two output numbers. E is incorrect because there was a correct answer.
Self Test Answers
6. D. 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. A, B, C, and E, are incorrect based on the program logic described above. 7. A, B, C, and D. A is correct because when a floating-point number (a double in this case) is cast to an int, it simply loses the digits after the decimal. B and D are correct because a long can be cast into a byte. If the long is over 127, it loses its most significant (leftmost) bits. C actually works, even though a cast is not necessary, because a long can store a byte. There are no incorrect answer choices.
Logical Operators (Sun Objective 5.3)
8. B. 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. A, C, D, E, and F are incorrect based on the program logic described above. 9. C. In the first two iterations x is incremented once and y is not because of the short circuit && operator. In the third and forth iterations x and y are each incremented, and in the fifth iteration x is doubly incremented and y is incremented. A, B, D, E, and F are incorrect based on the program logic described above. 10. B. The & operator has a higher precedence than the | operator so that on line 6 b1 and b2 are evaluated together as are b2 & b3. The final b1 in line 8 is what causes that if test to be true. A, C, and D are incorrect based on the program logic described above. 11. B. This is an example of a nested ternary operator. The second evaluation (x < 22) is true, so the tiny value is assigned to sup. A, C, and D are incorrect based on the program logic described above. 12. C. The reference variables b and x both refer to the same boolean array. Count is incremented for each call to the set() method, and once again when the first if test is true. Because of the && short circuit operator, count is not incremented during the second if test. A, B, D, E, and F are incorrect based on the program logic described above.
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