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- Question
What will be the output of the program?
public class SqrtExample { public static void main(String [] args) { double value = -9.0; System.out.println( Math.sqrt(value)); } }
Options- A. 3.0
- B. -3.0
- C. NaN
- D. Compilation fails.
- Correct Answer
- NaN
ExplanationThe sqrt() method returns NaN (not a number) when it's argument is less than zero.
Java.lang Class problems
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- 1. What will be the output of the program?
String a = "ABCD"; String b = a.toLowerCase(); b.replace('a','d'); b.replace('b','c'); System.out.println(b);
Options- A. abcd
- B. ABCD
- C. dccd
- D. dcba Discuss
Correct Answer: abcd
Explanation:
String objects are immutable, they cannot be changed, in this case we are talking about the replace method which returns a new String object resulting from replacing all occurrences of oldChar in this string with newChar.b.replace(char oldChar, char newChar);
But since this is only a temporary String it must either be put to use straight away i.e.
System.out.println(b.replace('a','d'));
Or a new variable must be assigned its value i.e.
String c = b.replace('a','d');
- 2. What will be the output of the program?
public class BoolTest { public static void main(String [] args) { int result = 0; Boolean b1 = new Boolean("TRUE"); Boolean b2 = new Boolean("true"); Boolean b3 = new Boolean("tRuE"); Boolean b4 = new Boolean("false"); if (b1 == b2) /* Line 10 */ result = 1; if (b1.equals(b2) ) /* Line 12 */ result = result + 10; if (b2 == b4) /* Line 14 */ result = result + 100; if (b2.equals(b4) ) /* Line 16 */ result = result + 1000; if (b2.equals(b3) ) /* Line 18 */ result = result + 10000; System.out.println("result = " + result); } }
Options- A. 0
- B. 1
- C. 10
- D. 10010 Discuss
Correct Answer: 10010
Explanation:
Line 10 fails because b1 and b2 are two different objects. Lines 12 and 18 succeed because the Boolean String constructors are case insensitive. Lines 14 and 16 fail because true is not equal to false.- 3. What will be the output of the program?
String a = "newspaper"; a = a.substring(5,7); char b = a.charAt(1); a = a + b; System.out.println(a);
Options- A. apa
- B. app
- C. apea
- D. apep Discuss
Correct Answer: app
Explanation:
Both substring() and charAt() methods are indexed with a zero-base, and substring() returns a String of length arg2 - arg1.- 4. What will be the output of the program?
public class Test { public static void main(String[] args) { final StringBuffer a = new StringBuffer(); final StringBuffer b = new StringBuffer(); new Thread() { public void run() { System.out.print(a.append("A")); synchronized(b) { System.out.print(b.append("B")); } } }.start(); new Thread() { public void run() { System.out.print(b.append("C")); synchronized(a) { System.out.print(a.append("D")); } } }.start(); } }
Options- A. ACCBAD
- B. ABBCAD
- C. CDDACB
- D. Indeterminate output Discuss
Correct Answer: Indeterminate output
Explanation:
It gives different output while executing the same compiled code at different times.- 5. What will be the output of the program?
public class ExamQuestion7 { static int j; static void methodA(int i) { boolean b; do { b = i<10 | methodB(4); /* Line 9 */ b = i<10 || methodB(8); /* Line 10 */ }while (!b); } static boolean methodB(int i) { j += i; return true; } public static void main(String[] args) { methodA(0); System.out.println( "j = " + j ); } }
Options- A. j = 0
- B. j = 4
- C. j = 8
- D. The code will run with no output Discuss
Correct Answer: j = 4
Explanation:
The lines to watch here are lines 9 & 10. Line 9 features the non-shortcut version of the OR operator so both of its operands will be evaluated and therefore methodB(4) is executed.However line 10 has the shortcut version of the OR operator and if the 1st of its operands evaluates to true (which in this case is true), then the 2nd operand isn't evaluated, so methodB(8) never gets called.
The loop is only executed once, b is initialized to false and is assigned true on line 9. Thus j = 4.
- 6. What will be the output of the program?
interface Foo141 { int k = 0; /* Line 3 */ } public class Test141 implements Foo141 { public static void main(String args[]) { int i; Test141 test141 = new Test141(); i = test141.k; /* Line 11 */ i = Test141.k; i = Foo141.k; } }
Options- A. Compilation fails.
- B. Compiles and runs ok.
- C. Compiles but throws an Exception at runtime.
- D. Compiles but throws a RuntimeException at runtime. Discuss
Correct Answer: Compiles and runs ok.
Explanation:
The variable k on line 3 is an interface constant, it is implicitly public, static, and final. Static variables can be referenced in two ways:Via a reference to any instance of the class (line 11)
Via the class name (line 12).
- 7. What will be the output of the program?
class Q207 { public static void main(String[] args) { int i1 = 5; int i2 = 6; String s1 = "7"; System.out.println(i1 + i2 + s1); /* Line 8 */ } }
Options- A. 18
- B. 117
- C. 567
- D. Compiler error Discuss
Correct Answer: 117
Explanation:
This question is about the + (plus) operator and the overriden + (string cocatanation) operator. The rules that apply when you have a mixed expression of numbers and strings are:If either operand is a String, the + operator concatenates the operands.
If both operands are numeric, the + operator adds the operands.
The expression on line 6 above can be read as "Add the values i1 and i2 together, then take the sum and convert it to a string and concatenate it with the String from the variable s1". In code, the compiler probably interprets the expression on line 8 above as:
System.out.println( new StringBuffer() .append(new Integer(i1 + i2).toString()) .append(s1) .toString() );- 8. What will be the output of the program?
String s = "hello"; Object o = s; if( o.equals(s) ) { System.out.println("A"); } else { System.out.println("B"); } if( s.equals(o) ) { System.out.println("C"); } else { System.out.println("D"); }- A
- B
- C
- D
Options- A. 1 and 3
- B. 2 and 4
- C. 3 and 4
- D. 1 and 2 Discuss
Correct Answer: 1 and 3
- 9. What will be the output of the program?
public class ObjComp { public static void main(String [] args ) { int result = 0; ObjComp oc = new ObjComp(); Object o = oc; if (o == oc) result = 1; if (o != oc) result = result + 10; if (o.equals(oc) ) result = result + 100; if (oc.equals(o) ) result = result + 1000; System.out.println("result = " + result); } }
Options- A. 1
- B. 10
- C. 101
- D. 1101 Discuss
Correct Answer: 1101
Explanation:
Even though o and oc are reference variables of different types, they are both referring to the same object. This means that == will resolve to true and that the default equals() method will also resolve to true.- 10. What will be the output of the program?
public class Example { public static void main(String [] args) { double values[] = {-2.3, -1.0, 0.25, 4}; int cnt = 0; for (int x=0; x < values.length; x++) { if (Math.round(values[x] + .5) == Math.ceil(values[x])) { ++cnt; } } System.out.println("same results " + cnt + " time(s)"); } }
Options- A. same results 0 time(s)
- B. same results 2 time(s)
- C. same results 4 time(s)
- D. Compilation fails. Discuss
Correct Answer: same results 2 time(s)
Explanation:
Math.round() adds .5 to the argument then performs a floor(). Since the code adds an additional .5 before round() is called, it's as if we are adding 1 then doing a floor(). The values that start out as integer values will in effect be incremented by 1 on the round() side but not on the ceil() side, and the noninteger values will end up equal.
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- 1. What will be the output of the program?