The answer is B because the code fragment contains a syntax error that prevents it from compiling. The code fragment tries to catch a StockException in line 10, but the catch block does not have a parameter of type StockException. The catch block should have a parameter of type StockException, such as:
catch (StockException e) { // handle the exception }
This is required by the Java syntax for the catch clause, which must have a parameter that is a subclass of Throwable. Without a parameter, the catch block is invalid and causes a compilation error.
Option A is incorrect because the program does not throw a StockException, as it does not compile.
Option C is incorrect because the program does not throw an OutofStockException, as it does not compile.
Option D is incorrect because the program does not throw a ClassCastException, as it does not compile. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
The try-with-resources Statement (The Java™ Tutorials > Essential Classes > Exceptions) The catch Blocks (The Java™ Tutorials > Essential Classes > Exceptions)

The code fragment will fail to compile because the play method in the Dog class is declared as private, which means that it cannot be accessed from outside the class. The main method is trying to call the play method on a Dog object, which is not allowed. Therefore, the code fragment will produce a compilation error.

Explanation
The code fragment will fail to compile because the play method in the Dog class is declared as private, which means that it cannot be accessed from outside the class. The main method is trying to call the play method on a Dog object, which is not allowed. Therefore, the code fragment will produce a compilation error.

Explanation
The answer is B because the code fragment contains a syntax error that prevents it from compiling. The code fragment tries to catch a StockException in line 10, but the catch block does not have a parameter of type StockException. The catch block should have a parameter of type StockException, such as:
catch (StockException e) { // handle the exception }
This is required by the Java syntax for the catch clause, which must have a parameter that is a subclass of Throwable. Without a parameter, the catch block is invalid and causes a compilation error.
Option A is incorrect because the program does not throw a StockException, as it does not compile.
Option C is incorrect because the program does not throw an OutofStockException, as it does not compile.
Option D is incorrect because the program does not throw a ClassCastException, as it does not compile.
References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
The try-with-resources Statement (The Java™ Tutorials > Essential Classes > Exceptions) The catch Blocks (The Java™ Tutorials > Essential Classes > Exceptions)

Explanation
The answer is B and E because the module-info file should include a provides directive and an exports directive to represent the service provider interface. The provides directive declares that the module provides an implementation of a service interface, which is com.transport.vehicle.cars.Car in this case. The with clause specifies the fully qualified name of the service provider class, which is com.transport.vehicle.cars.impl.CarImpl in this case. The exports directive declares that the module exports a package, which is com.transport.vehicle.cars in this case, to make it available to other modules. The package contains the service interface that other modules can use.
Option A is incorrect because requires is not the correct keyword to declare a service provider interface.
Requires declares that the module depends on another module, which is not the case here.
Option C is incorrect because it has a typo in the module name. It should be com.transport.vehicle.cars, not cm.transport.vehicle.cars.
Option D is incorrect because it has a typo in the keyword provides. It should be provides, not Provides. It also has a typo in the service interface name. It should be com.transport.vehicle.cars.Car, not com.transport.vehicle.cars.Car impl. It also has an unnecessary to clause, which is used to limit the accessibility of an exported package to specific modules.
Option F is incorrect because it exports the wrong package. It should export com.transport.vehicle.cars, not com.transport.vehicle.cars.impl. The impl package contains the service provider class, which should not be exposed to other modules.
Option G is incorrect because it exports the wrong package. It should export com.transport.vehicle.cars, not com.transport.vehicle. The vehicle package does not contain the service interface or the service provider class.
References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Java Modules – Service Interface Module – GeeksforGeeks
Java Service Provider Interface | Baeldung

Explanation
The code fragment will fail to compile because the parseInt method of the Integer class is a static method, which means that it can be invoked without creating an object of the class. However, the code is trying to invoke the parseInt method on an object of type Integer, which is not allowed. The correct way to invoke the parseInt method is by using the class name, such as Integer.parseInt (s). Therefore, the code fragment will produce a compilation error. References: Integer (Java SE 17 & JDK 17) – Oracle

Explanation
The answer is B because the code uses the writeObject and readObject methods of the ObjectOutputStream and ObjectInputStream classes to serialize and deserialize the Game object. These methods use the default serialization mechanism, which writes and reads the state of the object’s fields, including the inherited ones. Therefore, the title field of the Software class is also serialized and deserialized along with the players field of the Game class. The toString method of the Game class calls the toString method of the Software class using super.toString(), which returns the value of the title field.
Hence, when the deserialized object is printed, it shows “Software Game Software Game Chess 2”.
References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Serialization and Deserialization in Java with Example

Explanation
The code fragment is using the switch statement with the new Java 17 syntax. The switch statement checks the value of the variable rank and executes the corresponding case statement. In this case, the value of rank is 4, so the first case statement is executed, printing “Range1”. The second and third case statements are also executed, printing “Range2” and “Range3”. The default case statement is also executed, printing “Not a valid rank”. References: Java Language Changes – Oracle Help Center

Explanation
The code fragment in Option C invokes all callable objects in the workers set by using the ExecutorService’s invokeAll() method. This method takes a collection of Callable objects and returns a list of Future objects representing the results of the tasks. The other options are incorrect because they either use the wrong method (invokeAny() or submit()) or have syntax errors (missing parentheses or semicolons). References: AbstractExecutorService (Java SE 17 & JDK 17) – Oracle

Explanation
The code fragment uses the Collections.binarySearch method to search for the string “e3” in the list. The first search returns the index of the element, which is 2. The second search returns the index of the element, which is 0. The third search returns the index of the element, which is -4. The final result is
2. References: Collections (Java SE 17 & JDK 17) – Oracle

The answer is C because the code demonstrates the concept of method overloading and type conversion in Java. Method overloading allows different methods to have the same name but different parameters. Type conversion allows values of one data type to be assigned to another data type, either automatically or explicitly. In the code, the class Test has four methods named sum, each with different parameter types: int, float, and double. The main method creates an instance of Test and calls the sum method with different arguments. The compiler will choose the most specific method that matches the arguments, based on the following rules:
If there is an exact match between the argument types and the parameter types, that method is chosen.
If there is no exact match, but there is a method with compatible parameter types, that method is chosen. Compatible types are those that can be converted from one to another automatically, such as int to long or float to double.
If there is more than one method with compatible parameter types, the most specific method is chosen. The most specific method is the one whose parameter types are closest to the argument types in terms of size or precision.
In the code, the following method calls are made:
test.sum(10, 10.5) -> This matches the sum(int a, float b) method exactly, so it is chosen. The result is 20.5, which is converted to int and printed as 20 (B).
test.sum(10) -> This does not match any method exactly, but it matches the sum(double a) method with compatible types, as int can be converted to double automatically. The result is 10.0, which is printed as 10 (A).
test.sum(10.5, 10) -> This does not match any method exactly, but it matches two methods with compatible types: sum(float a, float b) and sum(double a, double b). The latter is more specific, as double is closer to the argument types than float. The result is 20.5, which is printed as 20 (D).
Therefore, the output is B A D. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Method Overloading in Java
Type conversion in Java with Examples
Java Method Overloading with automatic type conversions

Explanation
The code fragment is trying to obtain the java.io.Console object, which is a class that provides methods to access the character-based console device, if any, associated with the current Java virtual machine. The correct way to obtain the Console object is to call the static method Console console() in the java.lang.System class.
This method returns the unique Console object associated with the current Java virtual machine, if any.
Therefore, option A is correct, as it calls System.console() and assigns it to a Console variable. References:
https://docs.oracle.com/javase/17/docs/api/java.base/java/io/Console.html
https://docs.oracle.com/javase/17/docs/api/java.base/java/lang/System.html#console()
https://education.oracle.com/products/trackp_OCPJSE17
https://mylearn.oracle.com/ou/learning-path/java-se-17-developer/99487

Explanation
The code will not compile because the variable ‘x’ is declared as final and then it is being modified in the switch statement. This is not allowed in Java. A final variable is a variable whose value cannot be changed once it is initialized1. The switch statement tries to assign different values to ‘x’ depending on the value of ‘y’, which violates the final modifier. The compiler will report an error: The final local variable x cannot be assigned. It must be blank and not using a compound assignment. References: The final Keyword (The Java™ Tutorials > Learning the Java Language > Classes and Objects)

A module path is a sequence of directories that contain modules or JAR files. A named module is a module that has a name and a module descriptor (module-info.class) that declares its dependencies and exports. An automatic module is a module that does not have a module descriptor, but is derived from the name and contents of a JAR file. Both named and automatic modules can be placed on the module path, and they can be resolved by the Java runtime. An unnamed module is a special module that contains all the classes that are not in any other module, such as those on the class path. An unnamed module is not on the module path, but it can read all other modules.

Explanation
The answer is D because the code fragment reads the file “data.txt” and collects all the lines in the file into a single string, separated by hyphens. Then, it prints the resulting string. Next, it attempts to read the fourth line in the file (index 3) and print it. However, since the file only has three lines, an IndexOutOfBoundsException is thrown. References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Read contents of a file using Files class in Java

The answer is D because the code fragment reads the content of the in.txt file and writes it to the out.txt file. The content of the in.txt file is “23456789”. The code fragment uses a char array buffer of size 8 to read the content of the in.txt file. The while loop reads the content of the in.txt file and writes it to the out.txt file until the end of the file is reached. Therefore, the content of the out.txt file will be “0123456789”.

Explanation
The answer is C because the code fragment uses a different syntax and logic for the reduce operation than the other options. The reduce method in option C takes a single parameter, which is a BinaryOperator that combines two elements of the stream into one. The method returns an Optional, which may or may not contain a value depending on whether the stream is empty or not. The code fragment then adds 5 to the result of the reduce method, regardless of whether it is present or not. This may cause an exception if the Optional is empty, or produce a different value than the other options if the Optional is not empty.
The other options use a different syntax and logic for the reduce operation. They all take two parameters, which are an identity value and a BinaryOperator that combines an element of the stream with an accumulator.
The method returns the final accumulator value, which is equal to the identity value if the stream is empty, or the result of applying the BinaryOperator to all elements of the stream otherwise. The code fragments then add
5 to the result of the reduce method, which will always produce a valid value.
For example, suppose listOfNumbers contains [1, 2, 3]. Then, option A will perform the following steps:
Initialize accumulator to identity value 5
Apply BinaryOperator Integer::sum to accumulator and first element: 5 + 1 = 6 Update accumulator to 6 Apply BinaryOperator Integer::sum to accumulator and second element: 6 + 2 = 8 Update accumulator to 8 Apply BinaryOperator Integer::sum to accumulator and third element: 8 + 3 = 11 Update accumulator to 11 Return final accumulator value 11 Add 5 to final accumulator value: 11 + 5 = 16 Option B will perform the same steps as option A, except using a lambda expression instead of a method reference for the BinaryOperator. Option D will perform the same steps as option A, except using parallelStream instead of stream, which may change the order of applying the BinaryOperator but not the final result. Option E will perform the same steps as option A, except using identity value 0 instead of 5.
Option C, however, will perform the following steps:
Apply BinaryOperator Integer::sum to first and second element: 1 + 2 = 3 Apply BinaryOperator Integer::sum to previous result and third element: 3 + 3 = 6 Return Optional containing final result value 6 Add 5 to Optional value: Optional.of(6) + 5 = Optional.of(11) As you can see, option C produces a different value than the other options, and also uses a different syntax and logic for the reduce operation. References Oracle Certified Professional: Java SE 17 Developer Java SE 17 Developer OCP Oracle Certified Professional Java SE 17 Developer Study Guide Guide to Stream.reduce()

The code fragment is comparing the values of a, b, and c using the < and > operators. The first comparison, d, is checking if a is less than b and greater than c. Since a is equal to 2, b is equal to -2, and c is equal to -4, this comparison will evaluate to true. The second comparison, e, is checking if a is greater than b and a is greater than c. Since a is equal to 2, b is equal to -2, and c is equal to -4, this comparison will evaluate to false. Therefore, the result will be true 1 false 2. Reference: Operators (The Java™ Tutorials > Learning the Java Language – Oracle

Explanation
An automatic module name is derived from the name of the JAR file when it does not contain a module-info.class file. If the JAR file has an “Automatic-Module-Name” attribute in its main manifest, then its value is the module name. Otherwise, the module name is derived from the JAR file’s name by removing any version numbers and converting it to lower case. Therefore, for a JAR named display-ascii-0.2.jar, the automatic module name would be display-ascii, following these rules.

Explanation
The answer is E because the code fragment contains several syntax errors that prevent it from compiling.
Some of the errors are:
The enum declaration is missing a semicolon after the list of constants.
The enum constants are not capitalized, which violates the Java naming convention for enums.
The switch expression is missing parentheses around the variable name.
The case labels are missing colons after the enum constants.
The default label is missing a break statement, which causes a fall-through to the next case.
The println statement is missing a closing parenthesis and a semicolon.
A possible corrected version of the code fragment is:
enum Vehicle { BICYCLE, CAR, MOTORCYCLE, TRUCK; } public class Test { public static void main(String[] args) { Vehicle v = Vehicle.BICYCLE; switch (v) { case BICYCLE: System.out.print(“1”); break; case CAR: System.out.print(“3”); break; case MOTORCYCLE: System.out.print(“1”); break; case TRUCK: System.out.print(“2”); break; default: System.out.print(“0”); break; } System.out.println(); } } This would print 1 as the output. References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Enum Types
The switch Statement

Explanation
The tryLock () method of the Lock interface is a non-blocking attempt to acquire a lock. It returns true if the lock is available and acquired by the current thread, and false otherwise. It does not wait for the lock to be released by another thread. This is different from the lock () method, which blocks the current thread until the lock is acquired, and does not return any value. References:
https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/concurrent/locks/Lock.html#tryLock(), 3, 4