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

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. References:

• 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

The code fragment compares four pairs of strings using the equals() and intern() methods. The equals() method compares the content of two strings, while the intern() method returns a canonical representation of a string, which means that it returns a reference to an existing string with the same content in the string pool. The string pool is a memory area where strings are stored and reused to save space and improve performance. The results of the comparisons are as follows:

• s1.equals(s2): This returns true because both s1 and s2 have the same content, “Hello Java 17”.
• s1 == s2: This returns false because s1 and s2 are different objects with different references, even though they have the same content. The == operator compares the references of two objects, not their content.
• s1.intern() == s2.intern(): This returns true because both s1.intern() and s2.intern() return a reference to the same string object in the string pool, which has the content “Hello Java 17”. The intern() method ensures that there is only one copy of each distinct string value in the string pool.
• “Hello Java 17” == s2: This returns false because “Hello Java 17” is a string literal, which is automatically interned and stored in the string pool, while s2 is a string object created with the new operator, which is not interned by default and stored in the heap. Therefore, they have different references and are not equal using the == operator.

References: String (Java SE 17 & JDK 17) - Oracle

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)

The code fragment is using the bitwise operators & (AND), | (OR), and ^ (XOR) to perform operations on the binary representations of the integer values. The & operator returns a 1 in each bit position where both operands have a 1, the | operator returns a 1 in each bit position where either operand has a 1, and the ^ operator returns a 1 in each bit position where only one operand has a 1. The binary representations of the integer values are as follows:

• 1000 = 1111101000
• 100 = 1100100
• 101 = 1100101

The code fragment performs the following operations:

• x = x ^ y; // x becomes 1111010101, which is 1001 in decimal
• y = x ^ y; // y becomes 1100100, which is 100 in decimal
• x = x ^ y; // x becomes 1100101, which is 101 in decimal

The code fragment then prints out the values of x, y, and z, which are 1001, 100, and 1000 respectively. Therefore, option D is correct.

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.

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