1z0-830必殺問題集 & 1z0-830ウェブトレーニング

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Oracle 1z0-830 Exam Syllabus Topics:

SectionObjectives
Java Language Fundamentals- Java syntax and language structure
  • 1. Control flow statements
    • 2. Data types, variables, and operators
      Advanced Java Features (Java SE 21)- Modern language features
      • 1. Sealed classes
        • 2. Virtual threads (Project Loom)
          • 3. Pattern matching for switch
            • 4. Records and record patterns
              Exception Handling and Debugging- Error handling mechanisms
              • 1. Checked vs unchecked exceptions
                • 2. Try-with-resources
                  Concurrency and Multithreading- Thread management
                  • 1. Virtual threads and structured concurrency concepts
                    • 2. Thread lifecycle and synchronization
                      Database Connectivity (JDBC)- Database interaction
                      • 1. SQL execution and result handling
                        • 2. JDBC API usage
                          Object-Oriented Programming- Core OOP principles
                          • 1. Abstract classes and interfaces
                            • 2. Encapsulation, inheritance, polymorphism
                              Core APIs- Java standard library usage
                              • 1. Date and Time API
                                • 2. Collections Framework
                                  • 3. Streams and functional programming
                                    Input/Output and File Handling- NIO and file operations
                                    • 1. Serialization basics
                                      • 2. File, Path, and Streams APIs

                                        >> 1z0-830必殺問題集 <<

                                        最新の1z0-830必殺問題集 & 合格スムーズ1z0-830ウェブトレーニング | 一番優秀な1z0-830資格試験 Java SE 21 Developer Professional

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                                        Oracle Java SE 21 Developer Professional 認定 1z0-830 試験問題 (Q58-Q63):

                                        質問 # 58
                                        Given:
                                        java
                                        System.out.print(Boolean.logicalAnd(1 == 1, 2 < 1));
                                        System.out.print(Boolean.logicalOr(1 == 1, 2 < 1));
                                        System.out.print(Boolean.logicalXor(1 == 1, 2 < 1));
                                        What is printed?

                                        正解:E

                                        解説:
                                        In this code, three static methods from the Boolean class are used: logicalAnd, logicalOr, and logicalXor.
                                        Each method takes two boolean arguments and returns a boolean result based on the respective logical operation.
                                        Evaluation of Each Statement:
                                        * Boolean.logicalAnd(1 == 1, 2 < 1)
                                        * Operands:
                                        * 1 == 1 evaluates to true.
                                        * 2 < 1 evaluates to false.
                                        * Operation:
                                        * Boolean.logicalAnd(true, false) performs a logical AND operation.
                                        * The result is false because both operands must be true for the AND operation to return true.
                                        * Output:
                                        * System.out.print(false); prints false.
                                        * Boolean.logicalOr(1 == 1, 2 < 1)
                                        * Operands:
                                        * 1 == 1 evaluates to true.
                                        * 2 < 1 evaluates to false.
                                        * Operation:
                                        * Boolean.logicalOr(true, false) performs a logical OR operation.
                                        * The result is true because at least one operand is true.
                                        * Output:
                                        * System.out.print(true); prints true.
                                        * Boolean.logicalXor(1 == 1, 2 < 1)
                                        * Operands:
                                        * 1 == 1 evaluates to true.
                                        * 2 < 1 evaluates to false.
                                        * Operation:
                                        * Boolean.logicalXor(true, false) performs a logical XOR (exclusive OR) operation.
                                        * The result is true because exactly one operand is true.
                                        * Output:
                                        * System.out.print(true); prints true.
                                        Combined Output:
                                        Combining the outputs from each statement, the final printed result is:
                                        nginx
                                        falsetruetrue


                                        質問 # 59
                                        Given:
                                        java
                                        Object input = 42;
                                        String result = switch (input) {
                                        case String s -> "It's a string with value: " + s;
                                        case Double d -> "It's a double with value: " + d;
                                        case Integer i -> "It's an integer with value: " + i;
                                        };
                                        System.out.println(result);
                                        What is printed?

                                        正解:D

                                        解説:
                                        * Pattern Matching in switch
                                        * The switch expression introduced inJava 21supportspattern matchingfor different types.
                                        * However,a switch expression must be exhaustive, meaningit must cover all possible cases or provide a default case.
                                        * Why does compilation fail?
                                        * input is an Object, and the switch expression attempts to pattern-match it to String, Double, and Integer.
                                        * If input had been of another type (e.g., Float or Long), there would beno matching case, leading to anon-exhaustive switch.
                                        * Javarequires a default caseto ensure all possible inputs are covered.
                                        * Corrected Code (Adding a default Case)
                                        java
                                        Object input = 42;
                                        String result = switch (input) {
                                        case String s -> "It's a string with value: " + s;
                                        case Double d -> "It's a double with value: " + d;
                                        case Integer i -> "It's an integer with value: " + i;
                                        default -> "Unknown type";
                                        };
                                        System.out.println(result);
                                        * With this change, the codecompiles and runs successfully.
                                        * Output:
                                        vbnet
                                        It's an integer with value: 42
                                        Thus, the correct answer is:Compilation failsdue to a missing default case.
                                        References:
                                        * Java SE 21 - Pattern Matching for switch
                                        * Java SE 21 - switch Expressions


                                        質問 # 60
                                        Given:
                                        java
                                        Runnable task1 = () -> System.out.println("Executing Task-1");
                                        Callable<String> task2 = () -> {
                                        System.out.println("Executing Task-2");
                                        return "Task-2 Finish.";
                                        };
                                        ExecutorService execService = Executors.newCachedThreadPool();
                                        // INSERT CODE HERE
                                        execService.awaitTermination(3, TimeUnit.SECONDS);
                                        execService.shutdownNow();
                                        Which of the following statements, inserted in the code above, printsboth:
                                        "Executing Task-2" and "Executing Task-1"?

                                        正解:D、H

                                        解説:
                                        * Understanding ExecutorService Methods
                                        * execute(Runnable command)
                                        * Runs the task but only supports Runnable (not Callable).
                                        * #execService.execute(task2); fails because task2 is Callable<String>.
                                        * submit(Runnable task)
                                        * Submits a Runnable task for execution.
                                        * execService.submit(task1); executes "Executing Task-1".
                                        * submit(Callable<T> task)
                                        * Submits a Callable<T> task for execution.
                                        * execService.submit(task2); executes "Executing Task-2".
                                        * call() Does Not Exist in ExecutorService
                                        * #execService.call(task1); and execService.call(task2); are invalid.
                                        * run() Does Not Exist in ExecutorService
                                        * #execService.run(task1); and execService.run(task2); are invalid.
                                        * Correct Code to Print Both Messages:
                                        java
                                        execService.submit(task1);
                                        execService.submit(task2);
                                        Thus, the correct answer is:execService.submit(task1); execService.submit(task2); References:
                                        * Java SE 21 - ExecutorService
                                        * Java SE 21 - Callable and Runnable


                                        質問 # 61
                                        Which of the following statements oflocal variables declared with varareinvalid?(Choose 4)

                                        正解:A、D、E、F

                                        解説:
                                        1. Valid Use Cases of var
                                        * var is alocal variable type inferencefeature.
                                        * The compilerinfers the type from the assigned value.
                                        * Example of valid use:
                                        java
                                        var a = 10; // Type inferred as int
                                        var str = "Hello"; // Type inferred as String
                                        2. Analyzing the Given Statements
                                        Statement
                                        Valid/Invalid
                                        Reason
                                        var a = 1;
                                        Valid
                                        Type inferred as int.
                                        var b = 2, c = 3.0;
                                        #Invalid
                                        var doesnot allow multiple declarationsin one statement.
                                        var d[] = new int[4];
                                        #Invalid
                                        Array brackets []are not allowedwith var.
                                        var e;
                                        #Invalid
                                        varrequires an initializer(cannot be declared without assignment).
                                        var f = { 6 };
                                        #Invalid
                                        { 6 } is anarray initializer, which must have an explicit type.
                                        var h = (g = 7);
                                        Valid
                                        g is assigned 7, and h gets its value.
                                        Thus, the correct answers are:B, C, D, E
                                        References:
                                        * Java SE 21 - Local Variable Type Inference (var)
                                        * Java SE 21 - var Restrictions


                                        質問 # 62
                                        Given:
                                        java
                                        DoubleSummaryStatistics stats1 = new DoubleSummaryStatistics();
                                        stats1.accept(4.5);
                                        stats1.accept(6.0);
                                        DoubleSummaryStatistics stats2 = new DoubleSummaryStatistics();
                                        stats2.accept(3.0);
                                        stats2.accept(8.5);
                                        stats1.combine(stats2);
                                        System.out.println("Sum: " + stats1.getSum() + ", Max: " + stats1.getMax() + ", Avg: " + stats1.getAverage()); What is printed?

                                        正解:C

                                        解説:
                                        The DoubleSummaryStatistics class in Java is part of the java.util package and is used to collect and summarize statistics for a stream of double values. Let's analyze how the methods work:
                                        * Initialization and Data Insertion
                                        * stats1.accept(4.5); # Adds 4.5 to stats1.
                                        * stats1.accept(6.0); # Adds 6.0 to stats1.
                                        * stats2.accept(3.0); # Adds 3.0 to stats2.
                                        * stats2.accept(8.5); # Adds 8.5 to stats2.
                                        * Combining stats1 and stats2
                                        * stats1.combine(stats2); merges stats2 into stats1, resulting in one statistics summary containing all values {4.5, 6.0, 3.0, 8.5}.
                                        * Calculating Output Values
                                        * Sum= 4.5 + 6.0 + 3.0 + 8.5 = 22.0
                                        * Max= 8.5
                                        * Average= (22.0) / 4 = 5.5
                                        Thus, the output is:
                                        yaml
                                        Sum: 22.0, Max: 8.5, Avg: 5.5
                                        References:
                                        * Java SE 21 & JDK 21 - DoubleSummaryStatistics
                                        * Java SE 21 - Streams and Statistical Operations


                                        質問 # 63
                                        ......

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