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NEW QUESTION # 30
Which two code fragments compile?
- A.
- B.
- C.
- D.
- E.
Answer: B,C
Explanation:
The two code fragments that compile are B and E. These are the only ones that use the correct syntax for declaring and initializing a var variable. The var keyword is a reserved type name that allows the compiler to infer the type of the variable based on the initializer expression. However, the var variable must have an initializer, and the initializer must not be null or a lambda expression. Therefore, option A is invalid because it does not have an initializer, option C is invalid because it has a null initializer, and option D is invalid because it has a lambda expression as an initializer. Option B is valid because it has a String initializer, and option E is valid because it has an int initializer.
https://docs.oracle.com/en/java/javase/17/language/local-variable-type-inference.html
NEW QUESTION # 31
Given:
What is the result?
- A. Bicycle-1, car=3, motorcycle=1, truck=2)
- B. (Bicycle, car, motorcycle, truck)
- C. Bicycle =7, car=7, motorcycle=7, truck=7)
- D. Compilation fails.
- E. (3:bicycle, 0:car, 0motercycle, 5:truck)
Answer: D
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. Reference:
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
NEW QUESTION # 32
Given the directory structure:
Given the definition of the Doc class:
Which two are valid definition of the wordDoc class?
- A. Package p1;
Public final class WordDoc extends Doc () - B. Package p1, p2;
Public non-sealed class WordDoc extends Doc () - C. Package p1,
non-sealed abstract class WordDoc extends Doc () - D. Package p1;
Public non-sealed class wordDoc extends Doc () - E. Package p1;
Public class wordDoc extends Doc () - F. Package p1, p2;
Public sealed class WordDoc extends Doc ()
Answer: A,D
Explanation:
The correct answer is A and F because the wordDoc class must be a non-sealed class or a final class to extend the sealed Doc class. Option B is incorrect because the wordDoc class must be non-sealed or final. Option C is incorrect because the wordDoc class cannot be in a different package than the Doc class. Option D is incorrect because the wordDoc class cannot be a sealed class. Option E is incorrect because the wordDoc class cannot be an abstract class. Reference: Oracle Certified Professional: Java SE 17 Developer, 3 Sealed Classes - Oracle Help Center
NEW QUESTION # 33
Given the code fragment:
Which action sorts the book list?
- A. At Line n2, replace books,sort() with books.stream().sort(0.
- B. At Line n1, convert type to mutable array type.
- C. At line n1, convert books type to mutable ArrayList type.
- D. At Line n2, replace compareTo () with compare ().
Answer: D
Explanation:
Explanation
The code fragment is trying to sort a list of books using the Collections.sort() method. The correct answer is D, because the compareTo() method is not the correct way to compare two objects in a Comparator. The compare() method is the correct way to compare two objects in a Comparator and return an int value that indicates their order1. The compareTo() method is used to implement the Comparable interface, which defines the natural order of objects of a class2. The other options are incorrect because they either do not change the type of the list, which is already mutable, or they do not use the correct syntax for sorting a stream, which requires a terminal operation such as collect()3. References: Comparator (Java SE 17 & JDK 17), Comparable (Java SE 17 & JDK 17), Stream (Java SE 17 & JDK 17)
NEW QUESTION # 34
Given the code fragment:
What is the result?
- A. 0
- B. 1
- C. 2
- D. 3
- E. 4
- F. 5
Answer: B
Explanation:
The code fragment is creating a string variable "a" with the value "Hello! Java". Then, it is printing the index of "Java" in "a". Next, it is replacing "Hello!" with "Welcome!" in "a". Then, it is printing the index of "Java" in "a". Finally, it is creating a new StringBuilder object "b" with the value of "a" and printing the index of "Java" in "b". The output will be 8109 because the index of "Java" in "a" is 8, the index of "Java" in "a" after replacing "Hello!" with "Welcome!" is 10, and the index of "Java" in "b" is 9. Reference: Oracle Java SE 17 Developer source and documents: [String (Java SE 17 & JDK 17)], [StringBuilder (Java SE 17 & JDK 17)]
NEW QUESTION # 35
Given the code fragment:
What is the result?
- A. falase 0
true 1 - B. false 1
false 2 - C. true 1
false 2 - D. false 1
ture 2
Answer: C
Explanation:
Explanation
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. References: Operators (The Java™ Tutorials > Learning the Java Language - Oracle
NEW QUESTION # 36
Given:
What is the result?
- A. 1001
100
1000 - B. 1001
1001
1000 - C. 101
101
1000 - D. 100
100
1000
Answer: A
Explanation:
Explanation
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.
NEW QUESTION # 37
Given the code fragment:
Which code line n1, obtains the java.io.Console object?
- A.
- B.
- C.
- D.
- E.
Answer: A
Explanation:
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
NEW QUESTION # 38
Given:
What is the result?
- A. 1001
100
1000 - B. 1001
1001
1000 - C. 101
101
1000 - D. 100
100
1000
Answer: A
Explanation:
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.
NEW QUESTION # 39
Given:
Which action enables the code to compile?
- A. Replace 7 with public void display (string design) {
- B. Replace 2 with static string name;
- C. Replace 3 with private static void display () {
- D. Replace 15 with item.display (''Flower'');
Answer: A
Explanation:
The answer is C because the code fragment contains a syntax error in line 7, where the method display is declared without any parameter type. This causes a compilation error, as Java requires the parameter type to be specified for each method parameter. To fix this error, the parameter type should be added before the parameter name, such as string design. This will enable the code to compile and run without any errors. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Java Methods
NEW QUESTION # 40
Given the course table:
Given the code fragment:
- A. true
- B. 0
- C. false
- D. 1
Answer: A
Explanation:
Explanation
The code fragment will execute the update statement and set the course fee of the course with ID 1021 to
5000. The executeUpdate method returns an int value that indicates the number of rows affected by the SQL statement. In this case, only one row will be updated, so the result variable will be 1. The if statement will check if the result is greater than 0, which is true, and print "Updated successfully". Therefore, the output of the code fragment is true. References: https://education.oracle.com/products/trackp_OCPJSE17,
https://mylearn.oracle.com/ou/learning-path/java-se-17-developer/99487,
https://docs.oracle.com/en/java/javase/17/docs/api/java.sql/java/sql/Statement.html#executeUpdate(java.lang.Str
NEW QUESTION # 41
Given:
What is the result?
- A. Bicycle-1, car=3, motorcycle=1, truck=2)
- B. (Bicycle, car, motorcycle, truck)
- C. Bicycle =7, car=7, motorcycle=7, truck=7)
- D. Compilation fails.
- E. (3:bicycle, 0:car, 0motercycle, 5:truck)
Answer: D
Explanation:
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
NEW QUESTION # 42
Given the content of the in. tart file:
23456789
and the code fragment:
What is the content of the out .txt file?
- A. 012345678
- B. 01234567801234
- C. 012345678901234
- D. 0123456789234567
- E. 0123456789
- F. 01234567
Answer: E
Explanation:
Explanation
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".
NEW QUESTION # 43
Given:
What is the result?
- A. Software game write error
- B. Software Game Chess 2
- C. Software Game Software Game Chese 2
- D. Software Game Software Game chess 0
- E. Software Game Chess 0
- F. Software Game read error
Answer: C
Explanation:
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
NEW QUESTION # 44
Given the directory structure:
Given the definition of the Doc class:
Which two are valid definition of the wordDoc class?
- A. Package p1;
Public final class WordDoc extends Doc () - B. Package p1, p2;
Public non-sealed class WordDoc extends Doc () - C. Package p1,
non-sealed abstract class WordDoc extends Doc () - D. Package p1;
Public non-sealed class wordDoc extends Doc () - E. Package p1;
Public class wordDoc extends Doc () - F. Package p1, p2;
Public sealed class WordDoc extends Doc ()
Answer: A,D
Explanation:
Explanation
The correct answer is A and F because the wordDoc class must be a non-sealed class or a final class to extend the sealed Doc class. Option B is incorrect because the wordDoc class must be non-sealed or final. Option C is incorrect because the wordDoc class cannot be in a different package than the Doc class. Option D is incorrect because the wordDoc class cannot be a sealed class. Option E is incorrect because the wordDoc class cannot be an abstract class. References: Oracle Certified Professional: Java SE 17 Developer, 3 Sealed Classes - Oracle Help Center
NEW QUESTION # 45
Given:
What is the result?
- A. flipsflips
- B. runsruns
- C. runsflips
- D. flipsruns
- E. Compilation fails
Answer: E
Explanation:
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.
NEW QUESTION # 46
Given the code fragment:
Which action enables the code to compile?
- A. Replace record with void.
- B. Make the regNo variable public
- C. Make the regNo variable static.
- D. Replace thye regNo variable static
- E. Remove the regNO initialization statement.
Answer: B
Explanation:
The code will compile if the regNo variable is made public. This is because the regNo variable is being accessed in the main method of the App class, which is outside the scope of the Product class. Making the regNo variable public will allow it to be accessed from outside the class. Reference: https://education.oracle.com/products/trackp_OCPJSE17, https://mylearn.oracle.com/ou/learning-path/java-se-17-developer/99487, https://docs.oracle.com/javase/tutorial/java/javaOO/accesscontrol.html
NEW QUESTION # 47
Given:
What is the result?
- A. B A C
- B. B A D
- C. D D D
- D. D A D
Answer: B
Explanation:
Explanation
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
NEW QUESTION # 48
Given:
What is the result?
- A. A RuntimeException is thrown
- B. Tablechair Tablechair
- C. Wtablechair tableChair
- D. wTableChair TableChair
- E. Compilation fails
Answer: E
Explanation:
The code fragment will fail to compile because the class name and the constructor name do not match. The class name is Furniture, but the constructor name is Wtable. This will cause a syntax error. The correct way to define a constructor is to use the same name as the class name. Therefore, the code fragment should change the constructor name to Furniture or change the class name to Wtable.
NEW QUESTION # 49
Assume you have an automatic module from the module path display-ascii-0.2. jar. Which name is given to the automatic module based on the given JAR file?
- A. Display-ascii-0.2
- B. Display-ascii
- C. Display-ascii-0
- D. Display.ascii
Answer: B
Explanation:
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.
NEW QUESTION # 50
Given the code fragment:
What is the result?
- A. A NullPointerException is thrown
Cat
Cat - B. Cat
null - C. Cat
Dog - D. Dog
Dog
Answer: B
Explanation:
Explanation
The answer is E because the code fragment creates a new Pet object with the name "Dog" and assigns it to the variable p. Then, it assigns p to p1. Next, it changes the name of p1 to "Cat". Then, it assigns p1 to p. Finally, it sets p to null and prints the name of p and p1. The output will be "Cat" and "null" because p is set to null and p1 still points to the Pet object with the name "Cat".
NEW QUESTION # 51
......
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