Which of the following will directly stop the execution of a Thread?
A.
wait()
B.
notify()
C.
notifyall()
D.
exits synchronized code
Answer: Option A
Explanation:
Option A is correct. wait() causes the current thread to wait until another thread invokes the notify() method or the notifyAll() method for this object.
Option B is wrong. notify() - wakes up a single thread that is waiting on this object's monitor.
Option C is wrong. notifyAll() - wakes up all threads that are waiting on this object's monitor.
Option D is wrong. Typically, releasing a lock means the thread holding the lock (in other words, the thread currently in the synchronized method) exits the synchronized method. At that point, the lock is free until some other thread enters a synchronized method on that object. Does entering/exiting synchronized code mean that the thread execution stops? Not necessarily because the thread can still run code that is not synchronized. I think the word directly in the question gives us a clue. Exiting synchronized code does not directly stop the execution of a thread.
What allows the programmer to destroy an object x?
A.
x.delete()
B.
x.finalize()
C.
Runtime.getRuntime().gc()
D.
Only the garbage collection system can destroy an object.
Answer: Option D
Explanation:
Option D is correct. When an object is no longer referenced, it may be reclaimed by the garbage collector. If an object declares a finalizer, the finalizer is executed before the object is reclaimed to give the object a last chance to clean up resources that would not otherwise be released. When a class is no longer needed, it may be unloaded.
Option A is wrong. I found 4 delete() methods in all of the Java class structure. They are:
delete() - Method in class java.io.File : Deletes the file or directory denoted by this abstract pathname.
delete(int, int) - Method in class java.lang.StringBuffer : Removes the characters in a substring of this StringBuffer.
delete(int, int) - Method in interface javax.accessibility.AccessibleEditableText : Deletes the text between two indices
delete(int, int) - Method in class : javax.swing.text.JTextComponent.AccessibleJTextComponent; Deletes the text between two indices
None of these destroy the object to which they belong.
Option B is wrong. I found 19 finalize() methods. The most interesting, from this questions point of view, was the finalize() method in class java.lang.Object which is called by the garbage collector on an object when garbage collection determines that there are no more references to the object. This method does not destroy the object to which it belongs.
Option C is wrong. But it is interesting. The Runtime class has many methods, two of which are:
getRuntime() - Returns the runtime object associated with the current Java application.
gc() - Runs the garbage collector. Calling this method suggests that the Java virtual machine expend effort toward recycling unused objects in order to make the memory they currently occupy available for quick reuse. When control returns from the method call, the virtual machine has made its best effort to recycle all discarded objects. Interesting as this is, it doesn't destroy the object.
Switch statements are based on integer expressions and since both bytes and chars can implicitly be widened to an integer, these can also be used. Also shorts can be used. Short and Long are wrapper classes and reference types can not be used as variables.
public class X
{
public static void main(String [] args)
{
String names [] = new String[5];
for (int x=0; x < args.length; x++)
names[x] = args[x];
System.out.println(names[2]);
}
}
and the command line invocation is
> java X a b
A.
names
B.
null
C.
Compilation fails
D.
An exception is thrown at runtime
Answer: Option B
Explanation:
The names array is initialized with five null elements. Then elements 0 and 1 are assigned the String values "a" and "b" respectively (the command-line arguments passed to main). Elements of names array 2, 3, and 4 remain unassigned, so they have a value of null.
public class TestDogs
{
public static void main(String [] args)
{
Dog [][] theDogs = new Dog[3][];
System.out.println(theDogs[2][0].toString());
}
}
class Dog { }
A.
null
B.
theDogs
C.
Compilation fails
D.
An exception is thrown at runtime
Answer: Option D
Explanation:
The second dimension of the array referenced by theDogs has not been initialized. Attempting to access an uninitialized object element (System.out.println(theDogs[2][0].toString());) raises a NullPointerException.
An exception is thrown because in the code String s4 = args[4];, the array index (the fifth element) is out of bounds. The exception thrown is the cleverly named ArrayIndexOutOfBoundsException.
If assertions are compiled into a source file, and if no flags are included at runtime, assertions will execute by default.
B.
As of Java version 1.4, assertion statements are compiled by default.
C.
With the proper use of runtime arguments, it is possible to instruct the VM to disable assertions for a certain class, and to enable assertions for a certain package, at the same time.
D.
When evaluating command-line arguments, the VM gives -ea flags precedence over -da flags.
Answer: Option C
Explanation:
Option C is true because multiple VM flags can be used on a single invocation of a Java program.
Option A is incorrect because at runtime assertions are ignored by default.
Option B is incorrect because as of Java 1.4 you must add the argument -source 1.4 to the command line if you want the compiler to compile assertion statements.
Option D is incorrect because the VM evaluates all assertion flags left to right.
public class Test
{
public static void main (String[] args)
{
String foo = args[1];
String bar = args[2];
String baz = args[3];
System.out.println("baz = " + baz); /* Line 8 */
}
}
And the command line invocation:
> java Test red green blue
A.
baz =
B.
baz = null
C.
baz = blue
D.
Runtime Exception
Answer: Option D
Explanation:
When running the program you entered 3 arguments "red", "green" and "blue". When dealing with arrays in java you must remember ALL ARRAYS IN JAVA ARE ZERO BASED therefore args[0] becomes "red", args[1] becomes "green" and args[2] becomes "blue".
When the program entcounters line 8 above at runtime it looks for args[3] which has never been created therefore you get an
public static void main(String[] args)
{
Object obj = new Object()
{
public int hashCode()
{
return 42;
}
};
System.out.println(obj.hashCode());
}
A.
42
B.
Runtime Exception
C.
Compile Error at line 2
D.
Compile Error at line 5
Answer: Option A
Explanation:
This code is an example of an anonymous inner class. They can be declared to extend another class or implement a single interface. Since they have no name you can not use the "new" keyword on them.
In this case the annoynous class is extending the Object class. Within the {} you place the methods you want for that class. After this class has been declared its methods can be used by that object in the usual way e.g. objectname.annoymousClassMethod()
You want a class to have access to members of another class in the same package. Which is the most restrictive access that accomplishes this objective?
A.
public
B.
private
C.
protected
D.
default access
Answer: Option D
Explanation:
The only two real contenders are C and D. Protected access Option C makes a member accessible only to classes in the same package or subclass of the class. While default access Option D makes a member accessible only to classes in the same package.
Which three are valid method signatures in an interface?
private int getArea();
public float getVol(float x);
public void main(String [] args);
public static void main(String [] args);
boolean setFlag(Boolean [] test);
A.
1 and 2
B.
2, 3 and 5
C.
3, 4, and 5
D.
2 and 4
Answer: Option B
Explanation:
(2), (3), and (5). These are all valid interface method signatures.
(1), is incorrect because an interface method must be public; if it is not explicitly declared public it will be made public implicitly. (4) is incorrect because interface methods cannot be static.
Which is the valid declarations within an interface definition?
A.
public double methoda();
B.
public final double methoda();
C.
static void methoda(double d1);
D.
protected void methoda(double d1);
Answer: Option A
Explanation:
Option A is correct. A public access modifier is acceptable. The method prototypes in an interface are all abstract by virtue of their declaration, and should not be declared abstract.
Option B is wrong. The final modifier means that this method cannot be constructed in a subclass. A final method cannot be abstract.
Option C is wrong. static is concerned with the class and not an instance.
Option D is wrong. protected is not permitted when declaring a method of an interface. See information below.
Member declarations in an interface disallow the use of some declaration modifiers; you cannot use transient, volatile, or synchronized in a member declaration in an interface. Also, you may not use the private and protected specifiers when declaring members of an interface.
The default constructor initialises method variables.
The default constructor has the same access as its class.
The default constructor invokes the no-arg constructor of the superclass.
If a class lacks a no-arg constructor, the compiler always creates a default constructor.
The compiler creates a default constructor only when there are no other constructors for the class.
A.
1, 2 and 4
B.
2, 3 and 5
C.
3, 4 and 5
D.
1, 2 and 3
Answer: Option B
Explanation:
(2) sounds correct as in the example below
class CoffeeCup {
private int innerCoffee;
public CoffeeCup() {
}
public void add(int amount) {
innerCoffee += amount;
}
//...
}
The compiler gives default constructors the same access level as their class. In the example above, class CoffeeCup is public, so the default constructor is public. If CoffeeCup had been given package access, the default constructor would be given package access as well.
(3) is correct. The Java compiler generates at least one instance initialisation method for every class it compiles. In the Java class file, the instance initialisation method is named "<init>." For each constructor in the source code of a class, the Java compiler generates one <init>() method. If the class declares no constructors explicitly, the compiler generates a default no-arg constructor that just invokes the superclass's no-arg constructor. As with any other constructor, the compiler creates an <init>() method in the class file that corresponds to this default constructor.
(5) is correct. The compiler creates a default constructor if you do not declare any constructors in your class.
/* Missing statements ? */
public class NewTreeSet extends java.util.TreeSet
{
public static void main(String [] args)
{
java.util.TreeSet t = new java.util.TreeSet();
t.clear();
}
public void clear()
{
TreeMap m = new TreeMap();
m.clear();
}
}
which two statements, added independently at beginning of the program, allow the code to compile?
No statement is required
import java.util.*;
import.java.util.Tree*;
import java.util.TreeSet;
import java.util.TreeMap;
A.
1 only
B.
2 and 5
C.
3 and 4
D.
3 and 5
Answer: Option B
Explanation:
(2) and (5). TreeMap is the only class that must be imported. TreeSet does not need an import statement because it is described with a fully qualified name.
(1) is incorrect because TreeMap must be imported. (3) is incorrect syntax for an import statement. (4) is incorrect because it will not import TreeMap, which is required.
Statements (1), (2), (3), and (4) are correct. (1) 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.
(2) and (4) are correct because a long can be cast into a byte. If the long is over 127, it loses its most significant (leftmost) bits.
(3) actually works, even though a cast is not necessary, because a long can store a byte.
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);
}
}
A.
0
B.
1
C.
10
D.
10010
Answer: Option D
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.
public class Test138
{
public static void stringReplace (String text)
{
text = text.replace ('j' , 'c'); /* Line 5 */
}
public static void bufferReplace (StringBuffer text)
{
text = text.append ("c"); /* Line 9 */
}
public static void main (String args[])
{
String textString = new String ("java");
StringBuffer textBuffer = new StringBuffer ("java"); /* Line 14 */
stringReplace(textString);
bufferReplace(textBuffer);
System.out.println (textString + textBuffer);
}
}
A.
java
B.
javac
C.
javajavac
D.
Compile error
Answer: Option C
Explanation:
A string is immutable, it cannot be changed, that's the reason for the StringBuffer class. The stringReplace method does not change the string declared on line 14, so this remains set to "java".
Method parameters are always passed by value - a copy is passed into the method - if the copy changes, the original remains intact, line 5 changes the reference i.e. text points to a new String object, however this is lost when the method completes. The textBuffer is a StringBuffer so it can be changed.
This change is carried out on line 9, so "java" becomes "javac", the text reference on line 9 remains unchanged. This gives us the output of "javajavac"
class Boo
{
Boo(String s) { }
Boo() { }
}
class Bar extends Boo
{
Bar() { }
Bar(String s) {super(s);}
void zoo()
{
// insert code here
}
}
which one create an anonymous inner class from within class Bar?
A.
Boo f = new Boo(24) { };
B.
Boo f = new Bar() { };
C.
Bar f = new Boo(String s) { };
D.
Boo f = new Boo.Bar(String s) { };
Answer: Option B
Explanation:
Option B is correct because anonymous inner classes are no different from any other class when it comes to polymorphism. That means you are always allowed to declare a reference variable of the superclass type and have that reference variable refer to an instance of a subclass type, which in this case is an anonymous subclass of Bar. Since Bar is a subclass of Boo, it all works.
Option A is incorrect because it passes an int to the Boo constructor, and there is no matching constructor in the Boo class.
Option C is incorrect because it violates the rules of polymorphism—you cannot refer to a superclass type using a reference variable declared as the subclass type. The superclass is not guaranteed to have everything the subclass has.
Loading Test...
