Oracle Updated 1z0-830 Exam Questions and Answers by layan

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 Pérignon, ");

} catch (Exception e) {

System.out.print("Chablis, ");

} finally {

System.out.print("Saint-Émilion");

}

Since thrower() throws ArithmeticException, it is caught by catch (Exception e).

"Chablis, "is printed.

Thefinally block always executes, printing "Saint-Émilion".

Final Output

nginx

Beaujolais Nouveau, Chablis, Saint-Émilion

Thus, the correct answer is:Beaujolais Nouveau, Chablis, Saint-Émilion

References:

Java SE 21 - Exception Handling

Java SE 21 - finally Block Execution

Understanding Optional.ofNullable(null)

Optional.ofNullable(null); creates an empty Optional (i.e., it contains no value).

Optional.of(null); would throw a NullPointerException, but ofNullable(null); safely creates an empty Optional.

Execution of orElse, orElseGet, and orElseThrow

orElse("Baguette")

Since optionalName is empty, "Baguette" is returned.

bread = "Baguette"

Output:"bread:Baguette"

orElseGet(() -> "Frog legs")

Since optionalName is empty, "Frog legs" is returned from the lambda expression.

dish = "Frog legs"

Output:", dish:Frog legs"

orElseThrow(() -> new Exception())

Since optionalName is empty, an exception is thrown.

The catch block catches this exception and prints ", no cheese.".

Thus, the final output is:

makefile

bread:Baguette, dish:Frog legs, no cheese.

References:

Java SE 21 & JDK 21 - Optional

Java SE 21 - Functional Interfaces

In Java generics, wildcards are used to relax the type constraints of generic types. The extends wildcard () denotes an upper bounded wildcard, allowing any type that is a subclass of Type. Conversely, the super wildcard () denotes a lower bounded wildcard, allowing any type that is a superclass of Type.

Option A:

java

public List getListSuper() {

return new ArrayList();

}

Here, List represents a list that can hold IOException objects and objects of its supertypes. Since Exception is a superclass of IOException, ArrayList is compatible with List. Therefore, this method compiles successfully.

Option B:

java

public List getListExtends() {

return new ArrayList();

}

In this case, List represents a list that can hold objects of IOException and its subclasses. Since FileNotFoundException is a subclass of IOException, ArrayList is compatible with List. Thus, this method compiles successfully.

Option C:

java

public List getListExtends() {

return new ArrayList();

}

Here, List expects a list of IOException or its subclasses. However, Exception is a superclass of IOException, not a subclass. Therefore, ArrayList is not compatible with List, and this method will not compile.

Option D:

java

public List getListSuper() {

return new ArrayList();

}

In this scenario, List expects a list that can hold IOException objects and objects of its supertypes. Since FileNotFoundException is a subclass of IOException, ArrayList is not compatible with List, and this method will not compile.

Therefore, the methods in options A and B compile successfully, while those in options C and D do not.

Option A (Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();)

❌Compilation Fails

frenchAuthors is declared as List,notArrayList.

The correct way to declare a Map that allows storing List is to use List as the generic type,notArrayList.

Fix:

java

Map> authorsMap1 = new HashMap<>();

authorsMap1.put("FR", frenchAuthors);

Reason:The type ArrayList is more specific than List, and this would cause atype mismatcherror.

Option B (Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>>();)

❌Compilation Fails

? extends Listmakes the map read-onlyfor adding new elements.

The line authorsMap2.put("FR", frenchAuthors); causes acompilation errorbecause wildcard (? extends List) prevents modifying the map.

Fix:Remove the wildcard:

java

Map> 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 can be assigned to Map.

Using new HashMap>() with Map> isalloweddue to polymorphism.

Correct syntax:

java

Map> authorsMap4 = new HashMap>();

authorsMap4.put("FR", frenchAuthors);

Option E (Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>();)

Compiles Successfully

HashMap> isa valid instantiation.

Correct usage:

java

Map> 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