Online Java Programming Test - Java Programming Test - Random
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- Total number of questions: 20.
- Time allotted: 30 minutes.
- Each question carries 1 mark; there are no negative marks.
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- All the best!
Marks : 2/20
Test Review : View answers and explanation for this test.
Option B is the legal way to declare and initialize an array with five elements.
Option A is wrong because it shows an example of instantiating a class named Array, passing the integer value 5 to the object's constructor. If you don't see the brackets, you can be certain there is no actual array object! In other words, an Array object (instance of class Array) is not the same as an array object.
Option C is wrong because it shows a legal array declaration, but with no initialization.
Option D is wrong (and will not compile) because it declares an array with a size. Arrays must never be given a size when declared.
default access is the "package oriented" access modifier.
Option A and C are wrong because public and protected are less restrictive. Option B and D are wrong because abstract and synchronized are not access modifiers.
public class Outer
{
public void someOuterMethod()
{
//Line 5
}
public class Inner { }
public static void main(String[] argv)
{
Outer ot = new Outer();
//Line 10
}
}
Which of the following code fragments inserted, will allow to compile?
Option A compiles without problem.
Option B gives error - non-static variable cannot be referenced from a static context.
Option C package ot does not exist.
Option D gives error - non-static variable cannot be referenced from a static context.
class Super
{
public Integer getLength()
{
return new Integer(4);
}
}
public class Sub extends Super
{
public Long getLength()
{
return new Long(5);
}
public static void main(String[] args)
{
Super sooper = new Super();
Sub sub = new Sub();
System.out.println(
sooper.getLength().toString() + "," + sub.getLength().toString() );
}
}
Option D is correct, compilation fails - The return type of getLength( ) in the super class is an object of reference type Integer and the return type in the sub class is an object of reference type Long. In other words, it is not an override because of the change in the return type and it is also not an overload because the argument list has not changed.
class Test
{
public static void main(String [] args)
{
int x= 0;
int y= 0;
for (int z = 0; z < 5; z++)
{
if (( ++x > 2 ) && (++y > 2))
{
x++;
}
}
System.out.println(x + " " + y);
}
}
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.
class BitShift
{
public static void main(String [] args)
{
int x = 0x80000000;
System.out.print(x + " and ");
x = x >>> 31;
System.out.println(x);
}
}
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.
class PassS
{
public static void main(String [] args)
{
PassS p = new PassS();
p.start();
}
void start()
{
String s1 = "slip";
String s2 = fix(s1);
System.out.println(s1 + " " + s2);
}
String fix(String s1)
{
s1 = s1 + "stream";
System.out.print(s1 + " ");
return "stream";
}
}
When the fix() method is first entered, start()'s s1 and fix()'s s1 reference variables both refer to the same String object (with a value of "slip"). Fix()'s s1 is reassigned to a new object that is created when the concatenation occurs (this second String object has a value of "slipstream"). When the program returns to start(), another String object is created, referred to by s2 and with a value of "stream".
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");
}
}
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".
- 3/2
- 3<2
- 3*4
- 3<<2
(1) is wrong. 3/2 = 1 (integer arithmetic).
(2) is wrong. 3 < 2 = false.
(3) is correct. 3 * 4 = 12.
(4) is correct. 3 <<2= 12. In binary 3 is 11, now shift the bits two places to the left and we get 1100 which is 12 in binary (3*2*2).
int i = 1, j = -1;
switch (i)
{
case 0, 1: j = 1; /* Line 4 */
case 2: j = 2;
default: j = 0;
}
System.out.println("j = " + j);
Multi-constant case labels supported from Java 14, No compilation error and output will be j=0.
public class If1
{
static boolean b;
public static void main(String [] args)
{
short hand = 42;
if ( hand < 50 && !b ) /* Line 7 */
hand++;
if ( hand > 50 ); /* Line 9 */
else if ( hand > 40 )
{
hand += 7;
hand++;
}
else
--hand;
System.out.println(hand);
}
}
In Java, boolean instance variables are initialized to false, so the if test on line 7 is true and hand is incremented. Line 9 is legal syntax, a do nothing statement. The else-if is true so hand has 7 added to it and is then incremented.
public class Test
{
private static float[] f = new float[2];
public static void main (String[] args)
{
System.out.println("f[0] = " + f[0]);
}
}
The choices are between Option A and B, what this question is really testing is your knowledge of default values of an initialized array. This is an array type float i.e. it is a type that uses decimal point numbers therefore its initial value will be 0.0 and not 0
- The value returned by hashcode() is used in some collection classes to help locate objects.
- The hashcode() method is required to return a positive int value.
- The hashcode() method in the String class is the one inherited from Object.
- Two new empty String objects will produce identical hashcodes.
(2) is an incorrect statement because there is no such requirement.
(3) is an incorrect statement and therefore a correct answer because the hashcode for a string is computed from the characters in the string.
public class MyOuter
{
public static class MyInner
{
public static void foo() { }
}
}
MyInner is a static nested class, so it must be instantiated using the fully-scoped name of MyOuter.MyInner.
Option B is incorrect because it doesn't use the enclosing name in the new.
Option C is incorrect because it uses incorrect syntax. When you instantiate a nested class by invoking new on an instance of the enclosing class, you do not use the enclosing name. The difference between Option A and C is that Option C is calling new on an instance of the enclosing class rather than just new by itself.
Option D is incorrect because it doesn't use the enclosing class name in the variable declaration.
Option B is correct because in an interface all methods are abstract by default therefore they must be overridden by the implementing class. The Runnable interface only contains 1 method, the void run() method therefore it must be implemented.
Option A and D are incorrect because they are narrowing the access privileges i.e. package(default) access is narrower than public access.
Option C is not method in the Runnable interface therefore it is incorrect.
public class Test
{
public static int y;
public static void foo(int x)
{
System.out.print("foo ");
y = x;
}
public static int bar(int z)
{
System.out.print("bar ");
return y = z;
}
public static void main(String [] args )
{
int t = 0;
assert t > 0 : bar(7);
assert t > 1 : foo(8); /* Line 18 */
System.out.println("done ");
}
}
The foo() method returns void. It is a perfectly acceptable method, but because it returns void it cannot be used in an assert statement, so line 18 will not compile.
Option A is correct. The assertion status can be set for a named top-level class and any nested classes contained therein. This setting takes precedence over the class loader's default assertion status, and over any applicable per-package default. If the named class is not a top-level class, the change of status will have no effect on the actual assertion status of any class.
Option B is wrong. Is there such a thing as conditional compilation in Java?
Option C is wrong. For private methods - yes. But do not use assertions to check the parameters of a public method. An assert is inappropriate in public methods because the method guarantees that it will always enforce the argument checks. A public method must check its arguments whether or not assertions are enabled. Further, the assert construct does not throw an exception of the specified type. It can throw only an AssertionError.
Option D is wrong. Because you're never supposed to handle an assertion failure. That means don't catch it with a catch clause and attempt to recover.
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 */
}
}
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() );
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 );
}
}
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.
String d = "bookkeeper";
d.substring(1,7);
d = "w" + d;
d.append("woo"); /* Line 4 */
System.out.println(d);
In line 4 the code calls a StringBuffer method, append() on a String object.