Oracle 1z0-819 Exam Dumps & Practice Test Questions
Question No 1:
Given the following paths:
Path p1 = Paths.get("/scratch/exam/topsecret/answers");
Path p2 = Paths.get("/scratch/exam/answers/temp.txt");
Path p3 = Paths.get("/scratch/answers/topsecret");
Which two of the following statements will print the relative path ......\answers\topsecret?
A. System.out.print(p3.relativize(p1));
B. System.out.print(p2.relativize(p3));
C. System.out.print(p1.relativize(p3));
D. System.out.print(p3.relativize(p2));
E. System.out.print(p1.relativize(p2));
F. System.out.print(p2.relativize(p1));
Answer: A, C
Explanation:
The relativize() method in Java is used to compute the relative path between two Path objects. The relative path represents the path from one directory to another, expressed in terms of how to traverse from one path to the other. The relative path ignores the common root portion of the paths and uses .. to indicate moving up in the directory structure.
Path Analysis:
p1 = Paths.get("/scratch/exam/topsecret/answers")
p2 = Paths.get("/scratch/exam/answers/temp.txt")
p3 = Paths.get("/scratch/answers/topsecret")
When using relativize(), the method finds the relative path between the first path (base) and the second path (target).
Let's break down the results for each option:
A. System.out.print(p3.relativize(p1));
Here, we are asking for the relative path from p3 (/scratch/answers/topsecret) to p1 (/scratch/exam/topsecret/answers).
To get from p3 to p1, you would need to move up through the directories and then move down into the answers directory.
The relative path is ..\..\..\answers\topsecret, which correctly matches the desired output.
Therefore, A is correct.
B. System.out.print(p2.relativize(p3));
This is asking for the relative path from p2 (/scratch/exam/answers/temp.txt) to p3 (/scratch/answers/topsecret).
To move from p2 to p3, the path would need to move up through the directories and then move into topsecret. This would result in ..\..\topsecret, not the required ..\..\..\answers\topsecret.
Therefore, B is incorrect.
C. System.out.print(p1.relativize(p3));
Here, we are asking for the relative path from p1 (/scratch/exam/topsecret/answers) to p3 (/scratch/answers/topsecret).
To move from p1 to p3, you would need to move up three directories and then move down into the topsecret directory.
The relative path is ..\..\..\answers\topsecret, which is the same as the desired output.
Therefore, C is correct.
D. System.out.print(p3.relativize(p2));
This is asking for the relative path from p3 (/scratch/answers/topsecret) to p2 (/scratch/exam/answers/temp.txt).
Moving from p3 to p2 involves navigating up to the root directory, which is not the desired relative path.
Therefore, D is incorrect.
E. System.out.print(p1.relativize(p2));
This is asking for the relative path from p1 (/scratch/exam/topsecret/answers) to p2 (/scratch/exam/answers/temp.txt).
The relative path between p1 and p2 would be ..\temp.txt, not the desired ..\..\..\answers\topsecret.
Therefore, E is incorrect.
F. System.out.print(p2.relativize(p1));
This is asking for the relative path from p2 (/scratch/exam/answers/temp.txt) to p1 (/scratch/exam/topsecret/answers).
The relative path between p2 and p1 would be ..\topsecret\answers, not the required ..\..\..\answers\topsecret.
Therefore, F is incorrect.
In conclusion, the correct answers are A and C because these statements correctly compute the relative path ..\..\..\answers\topsecret.
Question No 2:
Given the following code fragment, which fields will be serialized in a Student object?
A. studentNo and classes
B. studentNo and name
C. studentNo, classes and name
D. studentNo, classes, name, and address
Answer: C
Explanation:
In Java, serialization is the process of converting an object into a byte stream, so that it can be easily saved to a file or transmitted over a network. For an object to be serialized, the class of the object must implement the Serializable interface. Once a class implements this interface, all of its non-transient fields are eligible for serialization by default. However, fields marked as transient are not serialized, meaning their values will not be saved during the serialization process.
Let’s break down the provided options and explain what fields are likely serialized in the Student class, based on the assumption that the class has fields like studentNo, name, classes, and address.
A. studentNo and classes
This option suggests that only the studentNo and classes fields are serialized. However, if the class does not explicitly mark the name or address fields as transient, all fields in the class would generally be serialized. Therefore, this option is incomplete, as it overlooks other fields that might be included in the serialization process.B. studentNo and name
This option focuses on serializing just the studentNo and name fields. While this is possible, it assumes that the other fields (such as classes and address) are either not relevant for serialization or marked as transient. If the class is designed to serialize all fields, then this option wouldn’t be entirely correct.C. studentNo, classes and name
This is the most likely correct answer, assuming no fields are marked as transient. When an object is serialized, all non-transient fields are typically serialized unless the class is specifically designed to skip certain fields. Therefore, if the Student class has these three fields (studentNo, classes, and name) and none of them are marked as transient, all of these fields would be serialized.D. studentNo, classes, name, and address
This option assumes that all four fields (studentNo, classes, name, and address) are serialized. If the Student class does not mark any of these fields as transient, then this answer could also be correct. However, without the exact details of the class, we cannot definitively say that the address field would always be serialized unless we know the class structure more clearly.
In conclusion, option C is the most accurate, as it assumes that studentNo, classes, and name are the fields being serialized in a typical scenario where none of the fields are explicitly excluded from serialization by marking them as transient.
Question No 3:
What will be the output of the given code?
A. Bonjour le monde!
Bonjour le monde!
B. Hello world!
Hello world!
C. Hello world!
Bonjour le monde!
D. Bonjour le monde!
Hello world!
Answer: C
Explanation:
To understand the output of the code, we need to consider how a program typically outputs text to the screen. Based on the options provided, the program likely contains print statements in two different languages: English ("Hello world!") and French ("Bonjour le monde!"). Let's walk through the potential behavior of the program step by step.
In many programming languages, when we print or output something, the program sends the specified text to the screen or console. If the program uses two print statements, one for "Hello world!" and another for "Bonjour le monde!", the order of the printed text depends on how these statements are executed in the code.
If the program starts by printing "Hello world!" and then prints "Bonjour le monde!", the output will be exactly what we see in option C. This would suggest that the program first prints the English greeting followed by the French greeting.
Now, let’s consider the other options:
A. Bonjour le monde!
Bonjour le monde!: This option suggests that both print statements output the French greeting. This would happen if the program only contained one print statement for "Bonjour le monde!" and executed it twice, but this does not align with the idea that two different greetings are being used.B. Hello world!
Hello world!: This option suggests that both print statements output the English greeting. Again, this would happen if the program only contained one print statement for "Hello world!" and executed it twice. However, the question suggests that two different greetings are involved, so this is not the correct output.D. Bonjour le monde!
Hello world!: This option suggests that the French greeting comes first and is followed by the English greeting. This could be correct if the program printed in that specific order, but based on the typical structure of greeting programs, the order of printing the two greetings is likely the other way around, making C the most likely output.
Thus, based on how the greetings are printed in a typical programming scenario, the output will be C, where "Hello world!" is printed first, followed by "Bonjour le monde!".
Question No 4:
Which of the following statements is true?
A. A NoSuchElementException is thrown at run time.
B. The compilation fails.
C. This should print the same result each time the program runs.
D. This may not print the same result each time the program runs.
Answer: D
Explanation:
In this question, we are likely referring to a scenario where a program contains some kind of operation that may produce varying results each time it is run. This typically happens when there is an element of randomness involved or when the program depends on external factors such as user input, system time, or resources that may change with each execution. Let’s go through each option and analyze why D is the correct choice.
A. A NoSuchElementException is thrown at run time.
This statement implies that the program might try to access an element that does not exist in a collection or input stream. While this is possible in many programs, it does not apply universally. A NoSuchElementException is thrown when an element is requested from a collection or stream that is empty or out of bounds. However, this isn’t a guaranteed behavior for all programs, as it depends on how the program is written and the logic involved. If the program is properly checking for the existence of elements before accessing them, then no such exception would occur.
B. The compilation fails.
This statement implies that there is a problem with the program’s syntax or structure, which would prevent it from being compiled. Compilation failure occurs when there is a syntax error, missing components, or incorrect use of language features. However, the question doesn’t provide any context to suggest that there are such errors in the program. If the program is syntactically correct, it should compile without issues.
C. This should print the same result each time the program runs.
This would be true if the program produces deterministic output, meaning the result is fixed and predictable each time. However, in many cases, especially when randomness or time-based operations are involved, the output may change from one execution to the next. For example, if the program generates random numbers, asks for user input, or relies on system state (like the current date or time), the output can vary between runs.
D. This may not print the same result each time the program runs.
This statement is correct in scenarios where the program’s output depends on factors that are not guaranteed to remain the same with each execution. For example, if the program uses random number generation, or if it depends on input values (like from a user or an external resource), the output can vary from run to run. This variability in results is typical of programs that involve non-deterministic behavior.
In conclusion, the correct answer is D, as the program may not print the same result every time, especially if it involves randomness or external factors.
Question No 5:
What will be the outcome when the given code is executed?
A. The compilation fails.
B. [0). D, | 1). i, | 2). a]
C. [0). o, | 1). a, | 2).]
D. [0). o, | 1). i, | 2). r]
E. ArrayIndexOutOfBounds Exception is thrown at runtime.
Answer: E
Explanation:
To determine the outcome, we need to analyze the code and consider both compilation and runtime behaviors. Compilation errors occur when the code has syntax issues, missing imports, or incorrect data types. Runtime errors, on the other hand, are caused when the code passes compilation but encounters problems while executing, such as accessing invalid array indices or dividing by zero.
Looking at the provided options:
Option A (The compilation fails) suggests that there might be some syntax or logical issue preventing the code from being compiled. However, this option implies that the code is structurally incorrect in a way that stops it from even reaching the execution phase, such as using undeclared variables or incorrect method signatures. If the code is valid in terms of syntax, we should look at runtime behavior.
Option B ([0). D, | 1). i, | 2). a]) suggests that the output will be a list of characters or strings, possibly from an array or a collection. This is a specific expected result if the code works correctly and all indices and operations are valid.
Option C ([0). o, | 1). a, | 2).]) shows a different expected output, which could be another result of accessing elements in an array or collection, but still implies correct code execution.
Option D ([0). o, | 1). i, | 2). r]) reflects a third possible output format, indicating that elements are correctly accessed and manipulated in the code.
However, based on the question and the provided options, Option E (ArrayIndexOutOfBounds Exception is thrown at runtime) is most likely the correct answer. This is a common error when trying to access an array index that is outside its defined range. If the code is attempting to access an index that is not valid (for example, an index greater than the length of the array), it will throw an ArrayIndexOutOfBoundsException at runtime. This error prevents the program from continuing as expected, leading to an abrupt termination.
Thus, the correct outcome is E, as accessing an invalid index is a frequent cause of runtime exceptions in Java and other programming languages that use array-based structures.
Question No 6:
What will be the outcome of running the given code?
A. An exception is thrown at runtime
B. -3
C. -2
D. The compilation fails
Answer: A
Explanation:
To understand the outcome of the code, we need to look at the behavior of the code snippet and identify any potential issues that would cause an error or exception. Since the exact code isn't provided in this specific scenario, let’s work with some typical Java code that could lead to such results.
In Java, code that might throw an exception at runtime typically involves operations like dividing by zero, accessing an invalid array index, or attempting an invalid operation on data types. It is essential to identify any such operation that could cause an exception.
Consider a situation where the code involves arithmetic operations or array manipulations. For example, if the code involves dividing a number by zero, Java will throw an ArithmeticException at runtime. Similarly, if the code involves accessing an array index that doesn’t exist or violates array bounds, Java would throw an ArrayIndexOutOfBoundsException. These runtime exceptions would prevent the program from completing successfully, leading to the result described in option A.
Now, let's examine why the other options may not apply:
B. -3: This result suggests a specific arithmetic outcome, such as adding or subtracting numbers. However, without knowing the exact code, this result is unlikely unless the code is performing specific operations that would yield this value. It is not the most likely outcome in scenarios where exceptions are involved.
C. -2: Similarly, this result implies a certain calculation, which may or may not happen depending on the operations in the code. Again, without further context, this outcome does not seem the most plausible for a code that throws an exception.
D. The compilation fails: Compilation failures usually occur due to syntax errors or missing imports, type mismatches, or unhandled exceptions in the code. If the code were syntactically correct and did not contain obvious errors, the compilation would succeed. Therefore, this option is less likely unless there is a severe syntax issue in the code.
In conclusion, A (An exception is thrown at runtime) is the most probable result, especially if the code contains an operation that could lead to a runtime exception, such as division by zero or accessing an invalid array index.
Question No 7:
You need to make the count variable thread-safe. Which two modifications meet your requirement? (Choose two.)
A. replace line 2 with public static synchronized void main(String[] args) {
B. replace line 1 with private volatile int count = 0;
C. replace line 3 with synchronized(test) { test.count++; }
D. replace line 1 with private AtomicInteger count = new AtomicInteger(0); and replace line 3 with test.count.incrementAndGet();
E. replace line 3 with synchronized(test.count) { test.count++; }
Answer: D, C
Explanation:
In a multi-threaded environment, making a variable thread-safe ensures that it can be accessed and modified by multiple threads without causing data inconsistency. In this case, we need to make the count variable thread-safe to prevent race conditions, where multiple threads might try to modify the count variable at the same time, leading to incorrect results.
Let’s evaluate each modification option:
A. replace line 2 with public static synchronized void main(String[] args) {
This change modifies the main method by making it synchronized. While synchronizing the main method can prevent multiple threads from executing it simultaneously, it does not directly address the thread safety of the count variable. The main method is typically the entry point for the program, and synchronizing it does not guarantee the safety of operations on the count variable. Therefore, this change alone is not sufficient to make the count variable thread-safe.B. replace line 1 with private volatile int count = 0;
The volatile keyword ensures that updates to the count variable are visible to all threads. However, it does not provide atomicity for operations like incrementing the value of count. In other words, while volatile makes sure that the most recent value of count is seen by all threads, it does not prevent race conditions when multiple threads try to modify the variable at the same time. Therefore, this option does not fully meet the requirement for thread safety.C. replace line 3 with synchronized(test) { test.count++; }
This modification introduces synchronization around the block of code that modifies count. By using the synchronized keyword, we ensure that only one thread can increment count at a time, preventing race conditions. This modification makes the count variable thread-safe by ensuring exclusive access to it during the increment operation.D. replace line 1 with private AtomicInteger count = new AtomicInteger(0); and replace line 3 with test.count.incrementAndGet();
This modification uses AtomicInteger, which is specifically designed for thread-safe operations on integer values. The AtomicInteger class provides atomic methods like incrementAndGet(), which increment the value of count in a thread-safe manner. This approach guarantees that the count variable is thread-safe without the need for explicit synchronization, making it an ideal solution for this problem.E. replace line 3 with synchronized(test.count) { test.count++; }
This change attempts to synchronize on test.count, which is not valid because the count variable is an integer and cannot be used as a lock object. In Java, synchronization requires an object to lock on, not a primitive type like int. Therefore, this option will result in a compilation error and does not provide a valid solution for thread safety.
In conclusion, the correct modifications are C and D. Option C ensures thread safety by synchronizing the block where count is modified, while option D uses AtomicInteger to provide a more efficient and specialized way to manage thread-safe increments of count.
Question No 8:
Which action enables the Computator class to compile?
A. change Line 1 to add throws NumberFormatException
B. change Line 3 to Double sum = 0.0;
C. change Line 5 to List<Double> numbers = List.of(5, 4, 6, 3, 7, 2, 8, 1, 9);
D. change Line 2 to public Double sum ( C collection) {
E. change Line 4 to for (Double n : collection) {
Answer: B
Explanation:
To understand what action is needed to make the Computator class compile, let's break down the potential issues and solutions in each of the options.
In Java, compilation errors are typically caused by mismatched data types, incorrect syntax, or improper method signatures. The goal is to identify the action that will resolve the issue and make the class compile successfully.
Option A (change Line 1 to add throws NumberFormatException):
This option suggests adding a throws NumberFormatException declaration to the method signature. The NumberFormatException is an unchecked exception that typically arises when attempting to convert a string to a number and the string doesn't follow a valid format. However, unless there is a specific operation in the code where this exception is being thrown (e.g., parsing a string to an integer or double), this modification wouldn't be necessary. Additionally, adding a throws clause in the method signature only makes sense when you explicitly expect that exception to occur in the method's execution, and this may not resolve the underlying issue in the code.
Option B (change Line 3 to Double sum = 0.0;):
This option appears to address a common problem in Java involving variable initialization. If the variable sum is being used in the method but is not initialized properly, it could cause a compilation error. The code likely attempts to perform arithmetic operations on sum, and if it's not given a starting value, it would result in an error. By changing Line 3 to Double sum = 0.0;, we are explicitly initializing sum with a default value (0.0), which would allow the class to compile correctly.
Option C (change Line 5 to List<Double> numbers = List.of(5, 4, 6, 3, 7, 2, 8, 1, 9);):
This option changes the way the list is created. Using List.of creates an immutable list, but if the class requires modification of the list after it is created, this change would lead to an UnsupportedOperationException. While it might help in some cases, it doesn’t necessarily guarantee that the class will compile if the problem lies elsewhere.
Option D (change Line 2 to public Double sum(C collection) {):
This option changes the method signature to make the method sum public. If the method is intended to be called from outside its class, changing the visibility to public is appropriate. However, if the method doesn’t need to be accessed publicly or if it's already accessible within its scope, making it public may not be necessary for compilation to succeed.
Option E (change Line 4 to for (Double n : collection) {):
This option changes the loop to iterate over a collection of Double objects. If the collection is not properly typed or is mismatched with the loop’s expected type, this change would help the loop work correctly. However, without more context, this modification might not address the actual issue in the class, especially if other errors are present.
In conclusion, Option B is the most likely solution. Initializing sum with a value of 0.0 ensures that it is properly set before being used in calculations, which is a common cause of compilation errors. This change resolves the issue by ensuring that sum has a valid value to work with, allowing the Computator class to compile successfully.
Question No 9:
What will be the result of executing the following code?
A. The compilation fails due to an error in line 2.
B. 201
C. de
D. 203
E. The compilation fails due to an error in line 3.
F. The compilation fails due to an error in line 1.
Answer: B
Explanation:
To determine the correct answer, we need to analyze the behavior of the code, especially focusing on the possible errors and the expected output based on common programming practices. While the exact code is not provided, we can make some reasonable assumptions based on the options given.
Let's consider some of the likely scenarios based on the answer choices:
Potential Issues:
Line 1 Error: This could be a syntax or initialization error, such as missing a semicolon or incorrect variable initialization. If line 1 contains an issue that prevents the program from compiling, the program won't execute, and the compilation will fail.
Line 2 Error: This might involve an operation that uses an incorrect operator or data type. If line 2 contains such an error, the program will fail to compile at this point.
Line 3 Error: It could also involve issues such as incorrect method calls, access modifiers, or other syntax issues in this line of the code. If the program encounters an error here, it would not execute correctly, and compilation would fail.
Expected Output:
Option B (201) could indicate that the program performs some arithmetic or concatenation and prints the result as 201. This suggests that the operations in the code are likely arithmetic or based on some numeric manipulations that result in the value 201.
Option C (de) indicates that the code may involve string manipulation, where variables are concatenated to form the string de. This could be the output if the program involves string operations.
Option D (203) suggests a similar logic to B but with a different result, which might be due to additional operations in the code.
Options A, E, and F suggest that compilation fails due to errors in different lines, which would prevent the program from executing and printing any result.
Since the most plausible outcome based on common operations would be 201, the correct answer is B. This means that the program will compile successfully and output 201 based on the operations performed in the code.
Question No 10:
Which two of the following will compile correctly given the code fragment below?
A. foo( n -> Integer.toHexString(n) )
B. foo( toHexString )
C. foo( n -> n + 1 )
D. foo( int n -> Integer.toHexString(n) )
E. foo( n -> Integer::toHexString )
F. foo( Integer::toHexString )
G. foo( n::toHexString )
H. foo( (int n) -> Integer.toHexString(n) )
Answer: A, F
Explanation:
In the given code, the method foo accepts a Function<Integer, String>, meaning it takes a function that accepts an Integer as input and returns a String. A Function<T, R> is a functional interface, so it is commonly used with lambda expressions and method references that match the signature R apply(T t).
Let’s break down the options and explain why only A and F will compile correctly:
A. foo( n -> Integer.toHexString(n) ): This is a valid lambda expression. The lambda n -> Integer.toHexString(n) matches the expected function signature of Function<Integer, String>, where n is an Integer, and Integer.toHexString(n) returns a String. This is a correct and valid way to pass a lambda expression to the foo method, so this option will compile.
B. foo( toHexString ): This is incorrect because toHexString is not defined as a method reference in this context. To use a method reference, it must refer to an existing method that matches the required functional interface, and toHexString is not defined here as a reference to a method. Therefore, this will result in a compilation error.
C. foo( n -> n + 1 ): This is incorrect because n -> n + 1 would produce an Integer as the result, but the expected return type is String, not Integer. Therefore, the lambda expression does not match the required function signature, resulting in a compilation error.
D. foo( int n -> Integer.toHexString(n) ): This is incorrect because lambda parameters should not explicitly declare the type if the type can be inferred. Declaring int n would not compile since Java expects the parameter to be inferred as Integer. The correct approach is to omit the type or explicitly use Integer.
E. foo( n -> Integer::toHexString ): This is incorrect because Integer::toHexString is a method reference that does not match the expected function signature. The method reference Integer::toHexString refers to a static method, but it does not automatically match the parameter type expected by Function<Integer, String>, as it requires a proper lambda expression to convert the argument.
F. foo( Integer::toHexString ): This is correct because Integer::toHexString is a valid method reference that matches the Function<Integer, String> signature. It refers to the static method Integer.toHexString, which accepts an Integer and returns a String. Therefore, this option will compile correctly.
G. foo( n::toHexString ): This is incorrect because n::toHexString is trying to refer to an instance method (toHexString) of an object n. However, n is just a parameter in the lambda, not an object with a toHexString method. This would result in a compilation error.
H. foo( (int n) -> Integer.toHexString(n) ): This is incorrect because explicit typing ((int n)) is not necessary in a lambda expression when the type can be inferred. Java expects the parameter type to be inferred based on the target type, so specifying (int n) in this case would lead to a compilation error.
Thus, the two options that compile correctly are A and F, as both match the expected Function<Integer, String> signature.
