1z1-829トレーニング最新認定問題をゲットOracle Java合格目指せ2024年08月27日 [Q30-Q54]

1z1-829トレーニング最新認定問題をゲットOracle Java合格目指せ2024年08月27日

認定トレーニング1z1-829試験問題集テストエンジン

Oracle 1Z0-829試験は、候補者のJava SE 17開発スキルを包括的にテストする試験です。試験は80問の多肢選択問題から構成され、制限時間は150分です。試験に合格するには、少なくとも63％以上のスコアを取得する必要があります。Javaクラスデザイン、Java I/Oの基礎、Java SE 17 APIの強化、Java SE 17の並列性の改善など、幅広いトピックが試験範囲に含まれます。この試験に合格すると、Java SE 17について深い理解を持ち、高品質なJavaアプリケーションを開発するためのスキルを身につけることができます。

 

質問 # 30
Which statement is true?

• A. The tryLock () method returns a boolean indicator immediately if it has managed to acquire the lock, otherwise it waits for the lock acquisition.
• B. The tryLock () method returns a boolean indicator immediately regardless if it has or has not managed to acquire the lock.
• C. The Lock () method returns a boolean indicator immediately regardless if it has or has not managed to acquire the lock
• D. The lock () method returns a boolean indicator immediately if it has managed to acquire the lock, otherwise it waits for the lock acquisition.

正解：B

解説：
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. Reference: https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/concurrent/locks/Lock.html#tryLock(), 3, 4, 5

質問 # 31
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
• B. Display.ascii
• C. Display-ascii
• D. Display-ascii-0.2

正解：C

解説：
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.

質問 # 32
Given:

What is the result?

• A. Software game write error
• B. Software Game Chess 2
• C. Software Game read error
• D. Software Game Software Game chess 0
• E. Software Game Chess 0
• F. Software Game Software Game Chese 2

正解：F

解説：
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". Reference:
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

質問 # 33
Given:

Which two should the module-info file include for it to represent the service provider interface?

• A. exports com.transport.vehicle.cars.Car;
• B. Requires cm.transport.vehicle,cars:
• C. Requires cm.transport.vehicle,cars:
• D. Provides.com.transport.vehicle.cars.Car with com.transport.vehicle.cars. impt, CatImpI;
• E. Provides.com.transport.vehicle.cars.Car impl,CarImp1 to com.transport.vehicle.cars. Cars
• F. Exports com.transport.vehicle.cars;
• G. Exports com.transport.vehicle;

正解：A、D

解説：
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

質問 # 34
Given the code fragment:

What is the result?

• A. Range 1
  Range 2
  Range 3
  Range 1
  Not a valida rank
• B. Range 1
• C. Range 1
  Range 2
  Range 3
• D. Range1
  Note a valid rank.

正解：A

解説：
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". Reference: Java Language Changes - Oracle Help Center

質問 # 35
Given the directory structure:

Given the definition of the Doc class:

Which two are valid definition of the wordDoc class?

• A. Package p1,
  non-sealed abstract class WordDoc extends Doc ()
• B. Package p1;
  Public class wordDoc extends Doc ()
• C. Package p1;
  Public non-sealed class wordDoc extends Doc ()
• D. Package p1, p2;
  Public sealed class WordDoc extends Doc ()
• E. Package p1;
  Public final class WordDoc extends Doc ()
• F. Package p1, p2;
  Public non-sealed class WordDoc extends Doc ()

正解：C、E

解説：
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

質問 # 36
Given:

and the code fragment:

• A. 300.00 CellPhone.ToyCar
• B. 100.00
  CellPhone,ToyCar,Motor,Fan
• C. 100.00 CellPhone,ToyCar
• D. 300.00
  CellPhone,ToyCar,Motor,Fan

正解：D

解説：
Explanation
The code fragment is using the Stream API to perform a reduction operation on a list of ElectricProduct objects. The reduction operation consists of three parts: an identity value, an accumulator function, and a combiner function. The identity value is the initial value of the result, which is 0.0 in this case. The accumulator function is a BiFunction that takes two arguments: the current result and the current element of the stream, and returns a new result. In this case, the accumulator function is (a,b) -> a + b.getPrice (), which means that it adds the price of each element to the current result. The combiner function is a BinaryOperator that takes two partial results and combines them into one. In this case, the combiner function is (a,b) -> a + b, which means that it adds the two partial results together.
The code fragment then applies a filter operation on the stream, which returns a new stream that contains only the elements that match the given predicate. The predicate is p -> p.getPrice () > 10, which means that it selects only the elements that have a price greater than 10. The code fragment then applies a map operation on the filtered stream, which returns a new stream that contains the results of applying the given function to each element. The function is p -> p.getName (), which means that it returns the name of each element.
The code fragment then calls the collect method on the mapped stream, which performs a mutable reduction operation on the elements of the stream using a Collector. The Collector is Collectors.joining (","), which means that it concatenates the elements of the stream into a single String, separated by commas.
The code fragment then prints out the result of the reduction operation and the result of the collect operation, separated by a new line. The result of the reduction operation is 300.00, which is the sum of the prices of all ElectricProduct objects that have a price greater than 10. The result of the collect operation is CellPhone,ToyCar,Motor,Fan, which is the concatenation of the names of all ElectricProduct objects that have a price greater than 10.
Therefore, the output of the code fragment is:
300.00 CellPhone,ToyCar,Motor,Fan
References: Stream (Java SE 17 & JDK 17) - Oracle, Collectors (Java SE 17 & JDK 17) - Oracle

質問 # 37
Given:

• A. Compilation fails
• B. there
• C. A NumberFormatException is thrown
• D. Hello

正解：A

解説：
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

質問 # 38
Given the code fragment:

What is the result:

• A. ABBCDE // the order of elements is unpredictable
• B. ABCE
• C. ABCDE // the order of elements is unpredictable
• D. ADEABCB // the order of element is unpredictable

正解：A

解説：
Explanation
The answer is D because the code fragment uses the Stream API to create two streams, s1 and s2, and then concatenates them using the concat() method. The resulting stream is then processed in parallel using the parallel() method, and the distinct() method is used to remove duplicate elements. Finally, the forEach() method is used to print the elements of the resulting stream to the console. Since the order of elements in a parallel stream is unpredictable, the output could be any of the options given, but option D is the most likely.
References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Parallelizing Streams

質問 # 39
Which statement is true?

• A. The tryLock () method returns a boolean indicator immediately if it has managed to acquire the lock, otherwise it waits for the lock acquisition.
• B. The tryLock () method returns a boolean indicator immediately regardless if it has or has not managed to acquire the lock.
• C. The Lock () method returns a boolean indicator immediately regardless if it has or has not managed to acquire the lock
• D. The lock () method returns a boolean indicator immediately if it has managed to acquire the lock, otherwise it waits for the lock acquisition.

正解：B

解説：
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

質問 # 40
Given the code fragment:

Which two statements at Line nl independently enable you to print 1250?

• A. Integer res= 250 + s2;
• B. Integer res = 250 + integer.parseint (s)
• C. Integer res = 250 + s:
• D. Integer res = 250 + integer . valueof (s);
• E. Integer res = 250 + integer (s2):
• F. Integer res = 250; Res = + s2;

正解：B、D

解説：
Explanation
The code fragment is creating a string variable "s" with the value "10_00" and an integer variable "s2" with the value 10. The string "s" is using an underscore as a separator for readability, which is allowed in Java SE
171. The question is asking for two statements that can add 250 to the numeric value of "s" and assign it to an integer variable "res". The correct answers are A and E because they use the methods parseInt and valueOf of the Integer class to convert the string "s" to an integer. Both methods interpret the string as a signed decimal integer and return the equivalent int or Integer value23. The other options are incorrect because they either use invalid syntax, such as B and C, or they do not convert the string "s" to an integer, such as D and F: References: Binary Literals (The Java™ Tutorials > Learning the Java Language > Numbers and Strings), Integer (Java SE 17 & JDK 17), Integer (Java SE 17 & JDK 17)

質問 # 41
Daylight Saving Time (DST) is the practice of advancing clocks at the start of spring by one hour and adjusting them backward by one hour in autumn.
Considering that in 2021, DST in Chicago (Illinois) ended on November 7th at 2 AM, and given the fragment:

What is the output?

• A. false
  true
• B. true
  false
• C. False
  false
• D. true
  true

正解：B

解説：
Explanation
The answer is A because the code fragment uses the ZoneId and ZonedDateTime classes to create two date-time objects with the same local date-time but different zone offsets. The ZoneId class represents a time-zone ID, such as America/Chicago, and the ZonedDateTime class represents a date-time with a time-zone in the ISO-8601 calendar system. The code fragment creates two ZonedDateTime objects with the same local date-time of 2021-11-07T01:30, but different zone IDs of America/Chicago and UTC. The code fragment then compares the two objects using the equals and isEqual methods.
The equals method compares the state of two objects for equality. In this case, it compares the local date-time, zone offset, and zone ID of the two ZonedDateTime objects. Since the zone offsets and zone IDs are different, the equals method returns false.
The isEqual method compares the instant of two temporal objects for equality. In this case, it compares the instant of the two ZonedDateTime objects, which is derived from the local date-time and zone offset. Since DST in Chicago ended on November 7th at 2 AM in 2021, the local date-time of 2021-11-07T01:30 in America/Chicago corresponds to the same instant as 2021-11-07T06:30 in UTC. Therefore, the isEqual method returns true.
Hence, the output is true false. References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
ZoneId (Java Platform SE 8 )
ZonedDateTime (Java Platform SE 8 )
Time Zone & Clock Changes in Chicago, Illinois, USA
Daylight Saving Time Changes 2023 in Chicago, USA

質問 # 42
Given:

What is the result?

• A. UserName
• B. User name (US)
• C. The program throws a MissingResourceException.
• D. User name (EN - US)
• E. User name (EN)

正解：C

解説：
The answer is B because the code fragment contains a logical error that causes a MissingResourceException at runtime. The code fragment tries to load a resource bundle with the base name "Captions.properties" and the locale "en_US". However, there is no such resource bundle available in the classpath. The available resource bundles are:
Captions.properties
Captions_en.properties
Captions_US.properties
Captions_en_US.properties
The ResourceBundle class follows a fallback mechanism to find the best matching resource bundle for a given locale. It first tries to find the resource bundle with the exact locale, then it tries to find the resource bundle with the same language and script, then it tries to find the resource bundle with the same language, and finally it tries to find the default resource bundle with no locale. If none of these resource bundles are found, it throws a MissingResourceException.
In this case, the code fragment is looking for a resource bundle with the base name "Captions.properties" and the locale "en_US". The ResourceBundle class will try to find the following resource bundles in order:
Captions.properties_en_US
Captions.properties_en
Captions.properties
However, none of these resource bundles exist in the classpath. Therefore, the ResourceBundle class will throw a MissingResourceException.
To fix this error, the code fragment should use the correct base name of the resource bundle family, which is "Captions" without the ".properties" extension. For example:
ResourceBundle captions = ResourceBundle.getBundle("Captions", currentLocale); This will load the appropriate resource bundle for the current locale, which is "Captions_en_US.properties" in this case. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
ResourceBundle (Java Platform SE 8 )
About the ResourceBundle Class (The Java™ Tutorials > Internationalization)

質問 # 43
Given the code fragment:

What is the result?

• A. Dog
  Dog
• B. Cat
  null
• C. A NullPointerException is thrown
  Cat
  Cat
• D. Cat
  Dog

正解：B

解説：
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".

質問 # 44
Given:

What is the result?

• A. 100
  100
  1000
• B. 1001
  1001
  1000
• C. 1001
  100
  1000
• D. 101
  101
  1000

正解：C

解説：
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.

質問 # 45
Given the code fragment:

What is the result?

• A. 1
  1
  1
• B. 0
  B 2
  C -2
• C. 2
  2
  0
• D. 1
  2
  -4

正解：C

解説：
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. Reference: Collections (Java SE 17 & JDK 17) - Oracle

質問 # 46
Which statement is true about modules?

• A. Only unnamed modules are on the module path.
• B. Automatic and named modules are on the module path.
• C. Automatic and unnamed modules are on the module path.
• D. Only automatic modules are on the module path.
• E. Only named modules are on the module path.

正解：B

解説：
Explanation
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.

質問 # 47
Given the code fragment:

Which code fragment returns different values?

• A. int sum = listOfNumbers. parallelStream () reduce ({m, n) -> m +n) orElse (5) +5;
• B. int sum = listOfNumbers. Stream () reduce (0, Integer:: sum) + 5
• C. int sum = listOfNumbers. Stream () reduce ( Integer:: sum) ; +5;
• D. int sum = listOfNumbers. Stream () reduce (5, (a, b) -> a + b) ;
• E. int sum = listOfNumbers. parallelStream () reduce (5, Integer:: sum) ;

正解：C

解説：
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()

質問 # 48
Given:

What is the result?

• A. D A D
• B. B A C
• C. D D D
• D. B A D

正解：D

解説：
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

質問 # 49
Which statement is true about modules?

• A. Only unnamed modules are on the module path.
• B. Automatic and named modules are on the module path.
• C. Automatic and unnamed modules are on the module path.
• D. Only automatic modules are on the module path.
• E. Only named modules are on the module path.

正解：B

解説：
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.

質問 # 50
Given the product class:

And the shop class:

What is the result?

• A. An exception is produced at runtime
• B. Cookie 0.0 2.99
• C. Cookie 0.0 0.0
• D. Compilation fails
• E. Cookie 3.99 2.99
• F. Cookie 2.99 2.99

正解：D

解説：
The code fragment will fail to compile because the readObject method in the Product class is missing the @Override annotation. The readObject method is a special method that is used to customize the deserialization process of an object. It must be declared as private, have no return type, and take a single parameter of type ObjectInputStream. It must also be annotated with @Override to indicate that it overrides the default behavior of the ObjectInputStream class. Without the @Override annotation, the compiler will treat the readObject method as a normal method and not as a deserialization hook. Therefore, the code fragment will produce a compilation error. Reference: Object Serialization - Oracle, [ObjectInputStream (Java SE 17 & JDK 17) - Oracle]

質問 # 51
Given:

What is the result?

• A. mA
• B. Mb
• C. MA
• D. Mb
  MC
• E. Mb

正解：A

解説：
Explanation
The code snippet is an example of Java SE 17 code. The code is checking if the object is an instance of class C and if it is, it will print "mC". If it is not an instance of class C, it will print "mA". In this case, the object is not an instance of class C, so the output will be "mA". References: Pattern Matching for instanceof - Oracle Help Center

質問 # 52
Given the code fragment:

What is the result:

• A. ABBCDE // the order of elements is unpredictable
• B. ABCE
• C. ABCDE // the order of elements is unpredictable
• D. ADEABCB // the order of element is unpredictable

正解：A

解説：
The answer is D because the code fragment uses the Stream API to create two streams, s1 and s2, and then concatenates them using the concat() method. The resulting stream is then processed in parallel using the parallel() method, and the distinct() method is used to remove duplicate elements. Finally, the forEach() method is used to print the elements of the resulting stream to the console. Since the order of elements in a parallel stream is unpredictable, the output could be any of the options given, but option D is the most likely. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Parallelizing Streams

質問 # 53
Given:

What is the result?

• A. 1 Snowy
• B. 0 Snowy
• C. Compilation fails
• D. 1 RAINY
• E. 0 CLOUDY

正解：B

解説：
Explanation
The code is defining an enum class called Forecast with three values: SUNNY, CLOUDY, and RAINY. The toString() method is overridden to always return "SNOWY". In the main method, the ordinal value of SUNNY is printed, which is 0, followed by the value of CLOUDY converted to uppercase, which is "CLOUDY".
However, since the toString() method of Forecast returns "SNOWY" regardless of the actual value, the output will be "0 SNOWY". References: Enum (Java SE 17 & JDK 17), Enum.EnumDesc (Java SE 17 & JDK 17)

質問 # 54
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