Correct Answers :

[B, C]

Explanation :

You can define multiple classes, interfaces, and enums in a Java source code file.
Option (a) is incorrect. The import statement applies to all the classes, interfaces, and enums defined within the same Java source code file.
Option (d) is incorrect. If a package statement is defined in the source code file, all of the classes, interfaces, and enums defined within it will exist in the same Java package.

Correct Answers :

[D]

Explanation :

The command-line arguments passed to the main method of a class do not contain the word Java and the name of the class.
Because the position of an array is zero-based, the method argument is assigned the following values:
args[0] -> one
args[1] -> two
args[2] -> three
args[3] -> four
The class prints two:three:four.

Correct Answers :

[C]

Explanation :

Option (a) is incorrect. This option defines a valid method but not a valid main method. The main method should be defined as a static method, which is missing from the method declaration in option (a).
Option (b) is incorrect. This option is similar to the method defined in option (a), with one difference. In this option, the square brackets are placed after the name of the method argument. The main method accepts an array as a method argument, and to define an array, the square brackets can be placed after either the data type or the method argument name.
Option (c) is correct. Extra spaces in a class are ignored by the Java compiler.
Option (d) is incorrect. The main method accepts an array of String as a method argument. The method in this option accepts a single String object.
Option (e) is incorrect. It isn’t a valid method definition and doesn’t specify the return type of the method. This line of code will not compile.

Correct Answers :

[B,C,D]

Explanation :

Only option (a) is a correct statement. Java primitive data types are predefined by the programming language. They can’t be defined by a programmer.

Correct Answers :

[A, C, D]

Explanation :

Option (a) is correct. The code at line 4 fails to compile because you can’t assign a negative value to a primitive char data type without casting.
Option (c) is correct. There is no primitive data type with the name "integer." The valid data types are int and Integer (a wrapper class with I in uppercase).
Option (d) is correct. The variable d remains undefined on line 7 because its declaration fails to compile on line 6. So the arithmetic expression (++a + b++ * c - d) that uses variable d fails to compile. There are no issues with using the variable c of the char data type in an arithmetic expression. The char data types are internally stored as unsigned integer values and can be used in arithmetic expressions.

Correct Answers :

[A]

Explanation :

The prefix increment operator (++) used with the variable a will increment its value before it is used in the expression ++a + b++ * c. The postfix increment operator (++) used with the variable b will increment its value after its initial value is used in the expression ++a + b++ * c.
Therefore, the expression ++a + b++ * c, evaluates with the following values: 11 + 20 * 30
Because the multiplication operator has a higher precedence than the addition operator, the values 20 and 30 are multiplied before the result is added to the value 11.
The example expression evaluates as follows:
(++a + b++ * c)
= 11 + 20 * 30
= 11 + 600
= 611

Correct Answers :

[A, B, C, D]

Explanation :

Option (e) is incorrect. There is no constraint on the number of arguments that can be passed on to a method, regardless of whether the method returns a value.
Option (f) is incorrect. You can’t return the value null for methods that return primitive data types. You can return null for methods that return objects (String is a class and not a primitive data type).

Correct Answers :

[B]

Explanation :

The class EJavaGuruPassObject1 defines two methods, someMethod and anotherMethod. The method someMethod modifies the value of the object parameter passed to it. Hence, the changes are visible within this method and in the calling method (method main). But the method anotherMethod reassigns the reference variable passed to it. Changes to any of the values of this object are limited to this method. They aren’t reflected in the calling method (the main method).

Correct Answers :

[C]

Explanation :

When primitive data types are passed to a method, the values of the variables in the calling method remain the same. This behavior doesn't depend on whether the primitive values are reassigned other values or modified by addition, subtraction, or multiplication—or any other operation.

Correct Answers :

[E]

Explanation :

The previous code won’t compile due to the following line of code:
arr[3] = longVar; This line of code tries to assign a value of type long to a variable of type int. Because Java does support implicit widening conversions for variables, the previous code fails to compile. Also, the previous code tries to trick you regarding your understanding of the following:
1) Assigning a char value to an int array element (arr[1] = c)
2) Adding a byte value to an int array element (arr[0] = b)
3) Whether an unassigned int array element is assigned a default value (arr[2])
4) Whether arr[0] + arr[1] + arr[2] + arr[3] prints the sum of all these values, or a concatenated value.

Correct Answers :

[E]

Explanation :

Because a char value can be assigned to an int value, you might assume that a char array can be assigned to an int array. But we’re talking about arrays of int and char primitives, which aren’t the same as a primitive int or char. Arrays themselves are reference variables, which refer to a collection of objects of similar type.

Correct Answers :

[A, D]

Explanation :

Option (b) is incorrect. This line of code won't compile because new array() isn't valid code. Unlike objects of other classes, an array isn't initialized using the keyword new followed by the word array. When the keyword new is used to initialize an array, it’s followed by the type of the array, not the word array.
Option (c) is incorrect.
To initialize a two-dimensional array, all of these values must be enclosed within another pair of curly braces, as shown in option (a).

Correct Answers :

[B]

Explanation :

The code has no compilation errors.
{
System.out.println("false");
}
The value false will print no matter what, regardless of whether the condition in the if construct evaluates to true or false.
Because the opening and closing braces for this code snippet are placed right after the if construct, it leads us to believe that this code snippet is the else part of the if construct. Also, note that an if construct uses the keyword else to define the else part. This keyword is missing in this question.
The if condition (that is, a++ > 10) evaluates to false because the postfix increment operator (a++) increments the value of the variable a immediately after its earlier value is used. 10 isn’t greater than 10 so this condition evaluates to false.

Correct Answers :

[D]

Explanation :

The code will fail to compile. The case labels require compile-time constant values, and the variable num2 doesn’t qualify as such. Although the variable num2 is defined as a final variable, it isn’t assigned a value with its declaration. The code assigns a literal value 20 to this variable after its declaration, but it isn’t considered to be a compile-time constant by the Java compiler.

Correct Answers :

[D]

Explanation :

The expressions used for both case labels—that is, 0 and 10*2-20—evaluate to the constant value 0. Because you can’t define duplicate case labels for the switch statement, the code will fail to compile with an error message that states that the code defines a duplicate case label.

Correct Answers :

[A]

Explanation :

Although the classes Cat and Rabbit seem to override the method jump, the class Cat doesn’t override the method jump() defined in the class Animal. The class Cat defines a method parameter with the method jump, which makes it an overloaded method, not an overridden method. Because the class Cat extends the class Animal, it has access to the following two overloaded jump methods:
void jump() { System.out.println("Animal"); }
void jump(int a) { System.out.println("Cat"); }
The following line of code creates an object of class Cat and assigns it to a variable of type Animal:
Animal cat = new Cat();
When you call the method jump on the previous object, it executes the method jump, which doesn’t accept any method parameters, printing the following value:
Animal
The following code will also print Animal and not Cat:
Cat cat = new Cat();
cat.jump();

Correct Answers :

[D]

Explanation :

Although the code seems to implement polymorphism using classes, note that neither of the classes Rose or Lily extends the class Flower. Hence, a variable of type Flower can’t be used to store objects of the classes Rose or Lily. The following lines of code will fail to compile:
Flower f1 = new Rose();
Flower f2 = new Lily();

Correct Answers :

[B]

Explanation :

Option (a) is incorrect. To implement polymorphism with classes, either an abstract class or a concrete class can be used as a base class.
Option (c) is incorrect. First of all, no code execution takes place at compile time.
Code can only execute at runtime. In polymorphism, the determination of the exact method to execute is deferred until runtime and is determined by the exact type of the object on which a method needs to be called.

Correct Answers :

[B]

Explanation :

No compilation errors exist with the code.
The method infibee throws StackOverflowError, which is not a checked exception. Even though your code should not throw an error, it is possible syntactically. Your code will compile successfully.
The call to the method infibee is immediately followed by the keyword return, which is supposed to end the execution of the method method. But the call to infibee is placed within a try-catch block, with a finally block. Because infibee doesn’t handle the error StackOverflowError itself, the control looks for the exception handler in the method method. This calling method doesn’t handle this error, but defines a finally block.
The control then executes the finally block. Because the code can’t find an appropriate handler to handle this error, it propagates to the JVM, which abruptly halts the code.

Correct Answers :

[B]

Explanation :

No direct relationship exists between exception handling and improved execution of code. Code that handles all the checked exceptions can throw unchecked exceptions and vice versa.