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NEW QUESTION # 15
Which two of the following aren't the correct ways to create a Stream?
- A. Stream<String> stream = Stream.builder().add("a").build();
- B. Stream stream = Stream.ofNullable("a");
- C. Stream stream = Stream.generate(() -> "a");
- D. Stream stream = Stream.of();
- E. Stream stream = Stream.of("a");
- F. Stream stream = new Stream();
- G. Stream stream = Stream.empty();
Answer: A,F
Explanation:
In Java, the Stream API provides several methods to create streams. However, not all approaches are valid.
NEW QUESTION # 16
Given:
java
interface SmartPhone {
boolean ring();
}
class Iphone15 implements SmartPhone {
boolean isRinging;
boolean ring() {
isRinging = !isRinging;
return isRinging;
}
}
Choose the right statement.
- A. Iphone15 class does not compile
- B. Everything compiles
- C. SmartPhone interface does not compile
- D. An exception is thrown at running Iphone15.ring();
Answer: A
Explanation:
In this code, the SmartPhone interface declares a method ring() with a boolean return type. The Iphone15 class implements the SmartPhone interface and provides an implementation for the ring() method.
However, in the Iphone15 class, the ring() method is declared without the public access modifier. In Java, when a class implements an interface, it must provide implementations for all the interface's methods with the same or a more accessible access level. Since interface methods are implicitly public, the implementing methods in the class must also be public. Failing to do so results in a compilation error.
Therefore, the Iphone15 class does not compile because the ring() method is not declared as public.
NEW QUESTION # 17
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. Rose
- C. Saint-Emilion
- D. Beaujolais Nouveau, Chablis, Dom Perignon, Saint-Emilion
Answer: A
Explanation:
* 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
NEW QUESTION # 18
Given:
java
public class OuterClass {
String outerField = "Outer field";
class InnerClass {
void accessMembers() {
System.out.println(outerField);
}
}
public static void main(String[] args) {
System.out.println("Inner class:");
System.out.println("------------");
OuterClass outerObject = new OuterClass();
InnerClass innerObject = new InnerClass(); // n1
innerObject.accessMembers(); // n2
}
}
What is printed?
- A. markdown
Inner class:
------------
Outer field - B. Nothing
- C. Compilation fails at line n1.
- D. An exception is thrown at runtime.
- E. Compilation fails at line n2.
Answer: C
Explanation:
* Understanding Inner Classes in Java
* Aninner class (non-static nested class)requires an instance of the outer classbefore it can be instantiated.
* Incorrect instantiationof the inner class at n1:
java
InnerClass innerObject = new InnerClass(); // Compilation error
* Since InnerClass is anon-staticinner class, itmust be created from an instance of OuterClass.
* Correct Way to Instantiate the Inner Class
java
OuterClass outerObject = new OuterClass();
OuterClass.InnerClass innerObject = outerObject.new InnerClass(); // Correct
* Thiscorrectly associatesthe inner class with an instance of OuterClass.
* Why Does Compilation Fail?
* The error occurs atline n1because InnerClass is beinginstantiated incorrectly.
Thus, the correct answer is:Compilation fails at line n1.
References:
* Java SE 21 - Nested and Inner Classes
* Java SE 21 - Accessing Outer Class Members
NEW QUESTION # 19
What does the following code print?
java
import java.util.stream.Stream;
public class StreamReduce {
public static void main(String[] args) {
Stream<String> stream = Stream.of("J", "a", "v", "a");
System.out.print(stream.reduce(String::concat));
}
}
- A. Compilation fails
- B. Optional[Java]
- C. Java
- D. null
Answer: B
Explanation:
In this code, a Stream of String elements is created containing the characters "J", "a", "v", and "a". The reduce method is then used with String::concat as the accumulator function.
The reduce method with a single BinaryOperator parameter performs a reduction on the elements of the stream, using an associative accumulation function, and returns an Optional describing the reduced value, if any. In this case, it concatenates the strings in the stream.
Since the stream contains elements, the reduction operation concatenates them to form the string "Java". The result is wrapped in an Optional, resulting in Optional[Java]. The print statement outputs this Optional object, displaying Optional[Java].
NEW QUESTION # 20
Given:
java
var array1 = new String[]{ "foo", "bar", "buz" };
var array2[] = { "foo", "bar", "buz" };
var array3 = new String[3] { "foo", "bar", "buz" };
var array4 = { "foo", "bar", "buz" };
String array5[] = new String[]{ "foo", "bar", "buz" };
Which arrays compile? (Select 2)
- A. array4
- B. array2
- C. array1
- D. array3
- E. array5
Answer: C,E
Explanation:
In Java, array initialization can be performed in several ways, but certain syntaxes are invalid and will cause compilation errors. Let's analyze each declaration:
* var array1 = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. The var keyword allows the compiler to infer the type from the initializer. Here, new String[]{ "foo", "bar", "buz" } creates an anonymous array of String with three elements. The compiler infers array1 as String[]. This syntax is correct and compiles successfully.
* var array2[] = { "foo", "bar", "buz" };
This declaration is invalid. While var can be used for type inference, appending [] after var is not allowed.
The correct syntax would be either String[] array2 = { "foo", "bar", "buz" }; or var array2 = new String[]{
"foo", "bar", "buz" };. Therefore, this line will cause a compilation error.
* var array3 = new String[3] { "foo", "bar", "buz" };
This declaration is invalid. In Java, when specifying the size of the array (new String[3]), you cannot simultaneously provide an initializer. The correct approach is either to provide the size without an initializer (new String[3]) or to provide the initializer without specifying the size (new String[]{ "foo", "bar", "buz" }).
Therefore, this line will cause a compilation error.
* var array4 = { "foo", "bar", "buz" };
This declaration is invalid. The array initializer { "foo", "bar", "buz" } can only be used in an array declaration when the type is explicitly provided. Since var relies on type inference and there's no explicit type provided here, this will cause a compilation error. The correct syntax would be String[] array4 = { "foo",
"bar", "buz" };.
* String array5[] = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. Here, String array5[] declares array5 as an array of String. The initializer new String[]{ "foo", "bar", "buz" } creates an array with three elements. This syntax is correct and compiles successfully.
Therefore, the declarations that compile successfully are array1 and array5.
References:
* Java SE 21 & JDK 21 - Local Variable Type Inference
* Java SE 21 & JDK 21 - Arrays
NEW QUESTION # 21
Given:
java
package com.vv;
import java.time.LocalDate;
public class FetchService {
public static void main(String[] args) throws Exception {
FetchService service = new FetchService();
String ack = service.fetch();
LocalDate date = service.fetch();
System.out.println(ack + " the " + date.toString());
}
public String fetch() {
return "ok";
}
public LocalDate fetch() {
return LocalDate.now();
}
}
What will be the output?
- A. ok the 2024-07-10
- B. ok the 2024-07-10T07:17:45.523939600
- C. Compilation fails
- D. An exception is thrown
Answer: C
Explanation:
In Java, method overloading allows multiple methods with the same name to exist in a class, provided they have different parameter lists (i.e., different number or types of parameters). However, having two methods with the exact same parameter list and only differing in return type is not permitted.
In the provided code, the FetchService class contains two fetch methods:
* public String fetch()
* public LocalDate fetch()
Both methods have identical parameter lists (none) but differ in their return types (String and LocalDate, respectively). This leads to a compilation error because the Java compiler cannot distinguish between the two methods based solely on return type.
The Java Language Specification (JLS) states:
"It is a compile-time error to declare two methods with override-equivalent signatures in a class." In this context, "override-equivalent" means that the methods have the same name and parameter types, regardless of their return types.
Therefore, the code will fail to compile due to the duplicate method signatures, and the correct answer is B:
Compilation fails.
NEW QUESTION # 22
Given:
java
List<Integer> integers = List.of(0, 1, 2);
integers.stream()
.peek(System.out::print)
.limit(2)
.forEach(i -> {});
What is the output of the given code fragment?
- A. 012
- B. Compilation fails
- C. Nothing
- D. 01
- E. An exception is thrown
Answer: D
Explanation:
In this code, a list of integers integers is created containing the elements 0, 1, and 2. A stream is then created from this list, and the following operations are performed in sequence:
* peek(System.out::print):
* The peek method is an intermediate operation that allows performing an action on each element as it is encountered in the stream. In this case, System.out::print is used to print each element.
However, since peek is intermediate, the printing occurs only when a terminal operation is executed.
* limit(2):
* The limit method is another intermediate operation that truncates the stream to contain no more than the specified number of elements. Here, it limits the stream to the first 2 elements.
* forEach(i -> {}):
* The forEach method is a terminal operation that performs the given action on each element of the stream. In this case, the action is an empty lambda expression (i -> {}), which does nothing for each element.
The sequence of operations can be visualized as follows:
* Original Stream Elements: 0, 1, 2
* After peek(System.out::print): Elements are printed as they are encountered.
* After limit(2): Stream is truncated to 0, 1.
* After forEach(i -> {}): No additional action; serves to trigger the processing.
Therefore, the output of the code is 01, corresponding to the first two elements of the list being printed due to the peek operation.
NEW QUESTION # 23
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. A NullPointerException is never thrown
- C. Only if assertions are enabled and the input argument isn't null
- D. Only if assertions are enabled and the input argument is null
- E. Only if assertions are disabled and the input argument isn't null
Answer: A
Explanation:
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
NEW QUESTION # 24
What is the output of the following snippet? (Assume the file exists)
java
Path path = Paths.get("C:\\home\\joe\\foo");
System.out.println(path.getName(0));
- A. C:
- B. Compilation error
- C. C
- D. IllegalArgumentException
- E. home
Answer: E
Explanation:
In Java's java.nio.file package, the Path class represents a file path in a file system. The Paths.get(String first, String... more) method is used to create a Path instance by converting a path string or URI.
In the provided code snippet, the Path object path is created with the string "C:\\home\\joe\\foo". This represents an absolute path on a Windows system.
The getName(int index) method of the Path class returns a name element of the path as a Path object. The index is zero-based, where index 0 corresponds to the first element in the path's name sequence. It's important to note that the root component (e.g., "C:\" on Windows) is not considered a name element and is not included in this sequence.
Therefore, for the path "C:\\home\\joe\\foo":
* Root Component:"C:\"
* Name Elements:
* Index 0: "home"
* Index 1: "joe"
* Index 2: "foo"
When path.getName(0) is called, it returns the first name element, which is "home". Thus, the output of the System.out.println statement is home.
NEW QUESTION # 25
Given:
java
String s = " ";
System.out.print("[" + s.strip());
s = " hello ";
System.out.print("," + s.strip());
s = "h i ";
System.out.print("," + s.strip() + "]");
What is printed?
- A. [ , hello ,hi ]
- B. [,hello,hi]
- C. [,hello,h i]
- D. [ ,hello,h i]
Answer: C
Explanation:
In this code, the strip() method is used to remove leading and trailing whitespace from strings. The strip() method, introduced in Java 11, is Unicode-aware and removes all leading and trailing characters that are considered whitespace according to the Unicode standard.
docs.oracle.com
Analysis of Each Statement:
* First Statement:
java
String s = " ";
System.out.print("[" + s.strip());
* The string s contains four spaces.
* Applying s.strip() removes all leading and trailing spaces, resulting in an empty string.
* The output is "[" followed by the empty string, so the printed result is "[".
* Second Statement:
java
s = " hello ";
System.out.print("," + s.strip());
* The string s is now " hello ".
* Applying s.strip() removes all leading and trailing spaces, resulting in "hello".
* The output is "," followed by "hello", so the printed result is ",hello".
* Third Statement:
java
s = "h i ";
System.out.print("," + s.strip() + "]");
* The string s is now "h i ".
* Applying s.strip() removes the trailing spaces, resulting in "h i".
* The output is "," followed by "h i" and then "]", so the printed result is ",h i]".
Combined Output:
Combining all parts, the final output is:
css
[,hello,h i]
NEW QUESTION # 26
Which of the following can be the body of a lambda expression?
- A. An expression and a statement
- B. None of the above
- C. Two statements
- D. A statement block
- E. Two expressions
Answer: D
Explanation:
In Java, a lambda expression can have two forms for its body:
* Single Expression:A concise form where the body consists of a single expression. The result of this expression is implicitly returned.
Example:
java
(a, b) -> a + b
In this example, (a, b) are the parameters, and a + b is the single expression that adds them together.
* Statement Block:A more detailed form where the body consists of a block of statements enclosed in braces {}. Within this block, you can have multiple statements, and if a return value is expected, you must explicitly use the return statement.
Example:
java
(a, b) -> {
int sum = a + b;
System.out.println("Sum is: " + sum);
return sum;
}
In this example, the lambda body is a statement block that performs multiple actions: it calculates the sum, prints it, and then returns the sum.
Given the options:
* A. Two statements:While a lambda body can contain multiple statements, they must be enclosed within a statement block {}. Simply having two statements without braces is not valid syntax for a lambda expression.
* B. An expression and a statement:Similar to option A, if a lambda body contains more than one element (be it expressions or statements), they need to be enclosed in a statement block.
* C. A statement block:This is correct. A lambda expression can have a body that is a statement block, allowing multiple statements enclosed in braces.
* D. None of the above:This is incorrect since option C is valid.
* E. Two expressions:As with options A and B, multiple expressions must be enclosed in a statement block to form a valid lambda body.
Therefore, the correct answer is C: A statement block.
NEW QUESTION # 27
Which three of the following are correct about the Java module system?
- A. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
- B. The unnamed module exports all of its packages.
- C. Code in an explicitly named module can access types in the unnamed module.
- D. The unnamed module can only access packages defined in the unnamed module.
- E. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
- F. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.
Answer: A,B,F
Explanation:
The Java Platform Module System (JPMS), introduced in Java 9, modularizes the Java platform and applications. Understanding the behavior of named and unnamed modules is crucial.
* B. The unnamed module exports all of its packages.
Correct. The unnamed module, which includes all code on the classpath, exports all of its packages. This means that any code can access the public types in these packages. However, the unnamed module cannot be explicitly required by named modules.
* C. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
Correct. In cases where a package is present in both a named module and the unnamed module, the version in the named module takes precedence. The package in the unnamed module is ignored to maintain module integrity and avoid conflicts.
* F. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.
Correct. When the module system cannot find a requested type in any known module, it defaults to searching the classpath (i.e., the unnamed module) to locate the type.
Incorrect Options:
* A. Code in an explicitly named module can access types in the unnamed module.
Incorrect. Named modules cannot access types in the unnamed module. The unnamed module can read from named modules, but the reverse is not allowed to ensure strong encapsulation.
* D. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
Incorrect. Adding a module descriptor (module-info.java) is not mandatory for applications developed before Java 9 to run on Java 11. Such applications can run in the unnamed module without modification.
* E. The unnamed module can only access packages defined in the unnamed module.
Incorrect. The unnamed module can access all packages exported by all named modules, in addition to its own packages.
NEW QUESTION # 28
Given:
java
List<String> frenchAuthors = new ArrayList<>();
frenchAuthors.add("Victor Hugo");
frenchAuthors.add("Gustave Flaubert");
Which compiles?
- A. Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>(); java authorsMap5.put("FR", frenchAuthors);
- B. Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();
java
authorsMap1.put("FR", frenchAuthors); - C. Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); java authorsMap4.put("FR", frenchAuthors);
- D. Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>> (); java authorsMap2.put("FR", frenchAuthors);
- E. var authorsMap3 = new HashMap<>();
java
authorsMap3.put("FR", frenchAuthors);
Answer: A,C,E
Explanation:
* 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
NEW QUESTION # 29
Given:
java
var counter = 0;
do {
System.out.print(counter + " ");
} while (++counter < 3);
What is printed?
- A. Compilation fails.
- B. 1 2 3 4
- C. 0 1 2
- D. 0 1 2 3
- E. 1 2 3
- F. An exception is thrown.
Answer: C
Explanation:
* Understanding do-while Execution
* A do-while loopexecutes at least oncebefore checking the condition.
* ++counter < 3 increments counterbeforeevaluating the condition.
* Step-by-Step Execution
* Iteration 1:counter = 0, print "0", then ++counter becomes 1, condition 1 < 3 istrue.
* Iteration 2:counter = 1, print "1", then ++counter becomes 2, condition 2 < 3 istrue.
* Iteration 3:counter = 2, print "2", then ++counter becomes 3, condition 3 < 3 isfalse, so loop exits.
* Final Output
0 1 2
Thus, the correct answer is:0 1 2
References:
* Java SE 21 - Control Flow Statements
* Java SE 21 - do-while Loop
NEW QUESTION # 30
......
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