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Dan Wells Dan Wells

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1z1-830資格問題対応、1z1-830無料試験

世界経済の急速な発展と国際的な競争の激化により、知識ベース経済の主要な地位は徐々に確立されています。多くの人が、良い仕事、1z1-830認定、より高い生活水準を求めています。良い仕事やより高い生活水準などを手に入れたいのであれば、変化する世界に歩調を合わせ、知識を更新することが非常に重要です。まず、適切な1z1-830クイズ準備を取得する必要があります。 1z1-830試験に合格して証明書を取得するだけなので、まともな仕事を得て、より多くのお金を稼ぐことができます。

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>> 1z1-830資格問題対応 <<

1z1-830無料試験 & 1z1-830対応資料

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Oracle Java SE 21 Developer Professional 認定 1z1-830 試験問題 (Q78-Q83):

質問 # 78
Given:
java
List<String> frenchAuthors = new ArrayList<>();
frenchAuthors.add("Victor Hugo");
frenchAuthors.add("Gustave Flaubert");
Which compiles?

A. var authorsMap3 = new HashMap<>();
java
authorsMap3.put("FR", frenchAuthors);
B. Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>> (); java authorsMap2.put("FR", frenchAuthors);
C. Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); java authorsMap4.put("FR", frenchAuthors);
D. Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>(); java authorsMap5.put("FR", frenchAuthors);
E. Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();
java
authorsMap1.put("FR", frenchAuthors);

正解：A、C、D

解説：
* Option A (Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();)
* #Compilation Fails
* frenchAuthors is declared as List<String>,notArrayList<String>.
* The correct way to declare a Map that allows storing List<String> is to use List<String> as the generic type,notArrayList<String>.
* Fix:
java
Map<String, List<String>> authorsMap1 = new HashMap<>();
authorsMap1.put("FR", frenchAuthors);
* Reason:The type ArrayList<String> is more specific than List<String>, and this would cause a type mismatcherror.
* Option B (Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>>();)
* #Compilation Fails
* ? extends List<String>makes the map read-onlyfor adding new elements.
* The line authorsMap2.put("FR", frenchAuthors); causes acompilation errorbecause wildcard (?
extends List<String>) prevents modifying the map.
* Fix:Remove the wildcard:
java
Map<String, List<String>> authorsMap2 = new HashMap<>();
authorsMap2.put("FR", frenchAuthors);
* Option C (var authorsMap3 = new HashMap<>();)
* Compiles Successfully
* The var keyword allows the compiler to infer the type.
* However,the inferred type is HashMap<Object, Object>, which may cause issues when retrieving values.
* Option D (Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>
>();)
* Compiles Successfully
* Valid declaration:HashMap<K, V> can be assigned to Map<K, V>.
* Using new HashMap<String, ArrayList<String>>() with Map<String, List<String>> isallowed due to polymorphism.
* Correct syntax:
java
Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); authorsMap4.put("FR", frenchAuthors);
* Option E (Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>();)
* Compiles Successfully
* HashMap<String, List<String>> isa valid instantiation.
* Correct usage:
java
Map<String, List<String>> authorsMap5 = new HashMap<>();
authorsMap5.put("FR", frenchAuthors);
Thus, the correct answers are:C, D, E
References:
* Java SE 21 - Generics and Type Inference
* Java SE 21 - var Keyword

質問 # 79
Given:
java
public class BoomBoom implements AutoCloseable {
public static void main(String[] args) {
try (BoomBoom boomBoom = new BoomBoom()) {
System.out.print("bim ");
throw new Exception();
} catch (Exception e) {
System.out.print("boom ");
}
}
@Override
public void close() throws Exception {
System.out.print("bam ");
throw new RuntimeException();
}
}
What is printed?

A. bim boom bam
B. bim bam boom
C. Compilation fails.
D. bim boom
E. bim bam followed by an exception

正解：B

解説：
* Understanding Try-With-Resources (AutoCloseable)
* BoomBoom implements AutoCloseable, meaning its close() method isautomatically calledat the end of the try block.
* Step-by-Step Execution
* Step 1: Enter Try Block
java
try (BoomBoom boomBoom = new BoomBoom()) {
System.out.print("bim ");
throw new Exception();
}
* "bim " is printed.
* Anexception (Exception) is thrown, butbefore it is handled, the close() method is executed.
* Step 2: close() is Called
java
@Override
public void close() throws Exception {
System.out.print("bam ");
throw new RuntimeException();
}
* "bam " is printed.
* A new RuntimeException is thrown, but it doesnot override the existing Exception yet.
* Step 3: Exception Handling
java
} catch (Exception e) {
System.out.print("boom ");
}
* The catch (Exception e)catches the original Exception from the try block.
* "boom " is printed.
* Final Output
nginx
bim bam boom
* Theoriginal Exception is caught, not the RuntimeException from close().
* TheRuntimeException from close() is ignoredbecause thecatch block is already handling Exception.
Thus, the correct answer is:bim bam boom
References:
* Java SE 21 - Try-With-Resources
* Java SE 21 - AutoCloseable Interface

質問 # 80
Given:
java
void verifyNotNull(Object input) {
boolean enabled = false;
assert enabled = true;
assert enabled;
System.out.println(input.toString());
assert input != null;
}
When does the given method throw a NullPointerException?

A. Only if assertions are disabled and the input argument is null
B. Only if assertions are enabled and the input argument is null
C. A NullPointerException is never thrown
D. Only if assertions are disabled and the input argument isn't null
E. Only if assertions are enabled and the input argument isn't null

正解：A

解説：
In the verifyNotNull method, the following operations are performed:
* Assertion to Enable Assertions:
java
boolean enabled = false;
assert enabled = true;
assert enabled;
* The variable enabled is initially set to false.
* The first assertion assert enabled = true; assigns true to enabled if assertions are enabled. If assertions are disabled, this assignment does not occur.
* The second assertion assert enabled; checks if enabled is true. If assertions are enabled and the previous assignment occurred, this assertion passes. If assertions are disabled, this assertion is ignored.
* Dereferencing the input Object:
java
System.out.println(input.toString());
* This line attempts to call the toString() method on the input object. If input is null, this will throw a NullPointerException.
* Assertion to Check input for null:
java
assert input != null;
* This assertion checks that input is not null. If input is null and assertions are enabled, this assertion will fail, throwing an AssertionError. If assertions are disabled, this assertion is ignored.
Analysis:
* If Assertions Are Enabled:
* The enabled variable is set to true by the first assertion, and the second assertion passes.
* If input is null, calling input.toString() will throw a NullPointerException before the final assertion is reached.
* If input is not null, input.toString() executes without issue, and the final assertion assert input != null; passes.
* If Assertions Are Disabled:
* The enabled variable remains false, but the assertions are ignored, so this has no effect.
* If input is null, calling input.toString() will throw a NullPointerException.
* If input is not null, input.toString() executes without issue.
Conclusion:
A NullPointerException is thrown if input is null, regardless of whether assertions are enabled or disabled.
Therefore, the correct answer is:
C: Only if assertions are disabled and the input argument is null

質問 # 81
You are working on a module named perfumery.shop that depends on another module named perfumery.
provider.
The perfumery.shop module should also make its package perfumery.shop.eaudeparfum available to other modules.
Which of the following is the correct file to declare the perfumery.shop module?

A. File name: module-info.perfumery.shop.java
java
module perfumery.shop {
requires perfumery.provider;
exports perfumery.shop.eaudeparfum.*;
}
B. File name: module-info.java
java
module perfumery.shop {
requires perfumery.provider;
exports perfumery.shop.eaudeparfum;
}
C. File name: module.java
java
module shop.perfumery {
requires perfumery.provider;
exports perfumery.shop.eaudeparfum;
}

正解：B

解説：
* Correct module descriptor file name
* A module declaration must be placed inside a file namedmodule-info.java.
* The incorrect filename module-info.perfumery.shop.javais invalid(Option A).
* The incorrect filename module.javais invalid(Option C).
* Correct module declaration
* The module declaration must match the name of the module (perfumery.shop).
* The requires perfumery.provider; directive specifies that perfumery.shop depends on perfumery.
provider.
* The exports perfumery.shop.eaudeparfum; statement allows the perfumery.shop.eaudeparfum package to beaccessible by other modules.
* The incorrect syntax exports perfumery.shop.eaudeparfum.*; in Option A isinvalid, as wildcards (*) arenot allowedin module exports.
Thus, the correct answer is:File name: module-info.java
References:
* Java SE 21 - Modules
* Java SE 21 - module-info.java File

質問 # 82
Given:
java
public class ExceptionPropagation {
public static void main(String[] args) {
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
}
static int thrower() {
try {
int i = 0;
return i / i;
} catch (NumberFormatException e) {
System.out.print("Rose");
return -1;
} finally {
System.out.print("Beaujolais Nouveau, ");
}
}
}
What is printed?

A. Beaujolais Nouveau, Chablis, Saint-Emilion
B. Beaujolais Nouveau, Chablis, Dom Perignon, Saint-Emilion
C. Rose
D. Saint-Emilion

正解：A

解説：
* Analyzing the thrower() Method Execution
java
int i = 0;
return i / i;
* i / i evaluates to 0 / 0, whichthrows ArithmeticException (/ by zero).
* Since catch (NumberFormatException e) doesnot matchArithmeticException, it is skipped.
* The finally block always executes, printing:
nginx
Beaujolais Nouveau,
* The exceptionpropagates backto main().
* Handling the Exception in main()
java
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
* Since thrower() throws ArithmeticException, it is caught by catch (Exception e).
* "Chablis, "is printed.
* Thefinally block always executes, printing "Saint-Emilion".
* Final Output
nginx
Beaujolais Nouveau, Chablis, Saint-Emilion
Thus, the correct answer is:Beaujolais Nouveau, Chablis, Saint-Emilion
References:
* Java SE 21 - Exception Handling
* Java SE 21 - finally Block Execution

質問 # 83
......

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我々の1z1-830試験問題集と回答は、より良いチャンスと良い人生のために、1z1-830実際試験に合格するために、あなたの助けになります、1z1-830実践教材は、あなたがそれを実現するのに役立ちます。

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