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- Question
What will be the output of the program?
String d = "bookkeeper"; d.substring(1,7); d = "w" + d; d.append("woo"); /* Line 4 */ System.out.println(d);
Options- A. wookkeewoo
- B. wbookkeeper
- C. wbookkeewoo
- D. Compilation fails.
- Correct Answer
- Compilation fails.
ExplanationIn line 4 the code calls a StringBuffer method, append() on a String object.
Java.lang Class problems
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- 1. What will be the output of the program?
int i = 1, j = 10; do { if(i++ > --j) /* Line 4 */ { continue; } } while (i < 5); System.out.println("i = " + i + "and j = " + j); /* Line 9 */
Options- A. i = 6 and j = 5
- B. i = 5 and j = 5
- C. i = 6 and j = 6
- D. i = 5 and j = 6 Discuss
Correct Answer: i = 5 and j = 6
Explanation:
This question is not testing your knowledge of the continue statement. It is testing your knowledge of the order of evaluation of operands. Basically the prefix and postfix unary operators have a higher order of evaluation than the relational operators. So on line 4 the variable i is incremented and the variable j is decremented before the greater than comparison is made. As the loop executes the comparison on line 4 will be:if(i > j)
if(2 > 9)
if(3 > 8)
if(4 > 7)
if(5 > 6) at this point i is not less than 5, therefore the loop terminates and line 9 outputs the values of i and j as 5 and 6 respectively.
The continue statement never gets to execute because i never reaches a value that is greater than j.
- 2. What will be the output of the program?
int i = (int) Math.random();
Options- A. i = 0
- B. i = 1
- C. value of i is undetermined
- D. Statement causes a compile error Discuss
Correct Answer: i = 0
Explanation:
Math.random() returns a double value greater than or equal to 0 and less than 1. Its value is stored to an int but as this is a narrowing conversion, a cast is needed to tell the compiler that you are aware that there may be a loss of precision.The value after the decimal point is lost when you cast a double to int and you are left with 0.
- 3. What will be the output of the program?
public class StringRef { public static void main(String [] args) { String s1 = "abc"; String s2 = "def"; String s3 = s2; /* Line 7 */ s2 = "ghi"; System.out.println(s1 + s2 + s3); } }
Options- A. abcdefghi
- B. abcdefdef
- C. abcghidef
- D. abcghighi Discuss
Correct Answer: abcghidef
Explanation:
After line 7 executes, both s2 and s3 refer to a String object that contains the value "def". When line 8 executes, a new String object is created with the value "ghi", to which s2 refers. The reference variable s3 still refers to the (immutable) String object with the value "def".- 4. What will be the output of the program?
String x = "xyz"; x.toUpperCase(); /* Line 2 */ String y = x.replace('Y', 'y'); y = y + "abc"; System.out.println(y);
Options- A. abcXyZ
- B. abcxyz
- C. xyzabc
- D. XyZabc Discuss
Correct Answer: xyzabc
Explanation:
Line 2 creates a new String object with the value "XYZ", but this new object is immediately lost because there is no reference to it. Line 3 creates a new String object referenced by y. This new String object has the value "xyz" because there was no "Y" in the String object referred to by x. Line 4 creates a new String object, appends "abc" to the value "xyz", and refers y to the result.- 5. What will be the output of the program?
public class WrapTest { public static void main(String [] args) { int result = 0; short s = 42; Long x = new Long("42"); Long y = new Long(42); Short z = new Short("42"); Short x2 = new Short(s); Integer y2 = new Integer("42"); Integer z2 = new Integer(42); if (x == y) /* Line 13 */ result = 1; if (x.equals(y) ) /* Line 15 */ result = result + 10; if (x.equals(z) ) /* Line 17 */ result = result + 100; if (x.equals(x2) ) /* Line 19 */ result = result + 1000; if (x.equals(z2) ) /* Line 21 */ result = result + 10000; System.out.println("result = " + result); } }
Options- A. result = 1
- B. result = 10
- C. result = 11
- D. result = 11010 Discuss
Correct Answer: result = 10
Explanation:
Line 13 fails because == compares reference values, not object values. Line 15 succeeds because both String and primitive wrapper constructors resolve to the same value (except for the Character wrapper). Lines 17, 19, and 21 fail because the equals() method fails if the object classes being compared are different and not in the same tree hierarchy.- 6. What will be the output of the program?
System.out.println(Math.sqrt(-4D));
Options- A. -2
- B. NaN
- C. Compile Error
- D. Runtime Exception Discuss
Correct Answer: NaN
Explanation:
It is not possible in regular mathematics to get a value for the square-root of a negative number therefore a NaN will be returned because the code is valid.- 7. What will be the output of the program?
public class NFE { public static void main(String [] args) { String s = "42"; try { s = s.concat(".5"); /* Line 8 */ double d = Double.parseDouble(s); s = Double.toString(d); int x = (int) Math.ceil(Double.valueOf(s).doubleValue()); System.out.println(x); } catch (NumberFormatException e) { System.out.println("bad number"); } } }
Options- A. 42
- B. 42.5
- C. 43
- D. bad number Discuss
Correct Answer: 43
Explanation:
All of this code is legal, and line 8 creates a new String with a value of "42.5". Lines 9 and 10 convert the String to a double and then back again. Line 11 is fun? Math.ceil()'s argument expression is evaluated first. We invoke the valueOf() method that returns an anonymous Double object (with a value of 42.5). Then the doubleValue() method is called (invoked on the newly created Double object), and returns a double primitive (there and back again), with a value of (you guessed it) 42.5. The ceil() method converts this to 43.0, which is cast to an int and assigned to x.- 8. What will be the output of the program?
String s = "ABC"; s.toLowerCase(); s += "def"; System.out.println(s);
Options- A. ABC
- B. abc
- C. ABCdef
- D. Compile Error Discuss
Correct Answer: ABCdef
Explanation:
String objects are immutable. The object s above is set to "ABC". Now ask yourself if this object is changed and if so where - remember strings are immutable.Line 2 returns a string object but does not change the originag string object s, so after line 2 s is still "ABC".
So what's happening on line 3? Java will treat line 3 like the following:
s = new StringBuffer().append(s).append("def").toString();
This effectively creates a new String object and stores its reference in the variable s, the old String object containing "ABC" is no longer referenced by a live thread and becomes available for garbage collection.
- 9. What will be the output of the program?
public class ExamQuestion6 { static int x; boolean catch() { x++; return true; } public static void main(String[] args) { x=0; if ((catch() | catch()) || catch()) x++; System.out.println(x); } }
Options- A. 1
- B. 2
- C. 3
- D. Compilation Fails Discuss
Correct Answer: Compilation Fails
Explanation:
Initially this looks like a question about the logical and logical shortcut operators "|" and "||" but on closer inspection it should be noticed that the name of the boolean method in this code is "catch". "catch" is a reserved keyword in the Java language and cannot be used as a method name. Hence Compilation will fail.- 10. What will be the output of the program?
try { Float f1 = new Float("3.0"); int x = f1.intValue(); byte b = f1.byteValue(); double d = f1.doubleValue(); System.out.println(x + b + d); } catch (NumberFormatException e) /* Line 9 */ { System.out.println("bad number"); /* Line 11 */ }
Options- A. 9.0
- B. bad number
- C. Compilation fails on line 9.
- D. Compilation fails on line 11. Discuss
Correct Answer: 9.0
Explanation:
The xxxValue() methods convert any numeric wrapper object's value to any primitive type. When narrowing is necessary, significant bits are dropped and the results are difficult to calculate.
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- 1. What will be the output of the program?