1Z0-811 Java Basics and Syntax Practice Test 7 — 15 Questions

Question 1

Which THREE are valid Java identifiers? (Choose three.)

Accepted Answer

$dollar. Option A ($dollar) is a valid Java identifier because the dollar sign ($) is a legal identifier-start character in Java, alongside letters, the underscore, and Unicode currency symbols. Option D (_value) is valid because an underscore is a legal identifier-start character, so a name beginning with _ followed by letters is accepted by the compiler. Option E (myVar123) is valid because it starts with a letter and continues with letters and digits, and digits are permitted in identifiers as long as they are not the first character. Option B (my var) is invalid because it contains a space, which is not a legal identifier character and would be parsed as two separate tokens. Option C (2ndValue) is invalid because it begins with the digit 2, and Java identifiers cannot start with a numeric digit.

Answer

my var

Answer

2ndValue

Question 2

What likely caused this compilation error?

Accepted Answer

The filename does not match the public class name. The compilation error occurs because Java requires that the public class name exactly matches the filename (including case) when the class is declared as public. If the filename is different from the public class name, the compiler will fail with an error indicating the class name is incorrect or cannot be found.

Answer

The main method is missing

Answer

The file is saved with a .txt extension instead of .java

Answer

The class name contains a typo

Question 3

A developer is working on a Java application that processes user input. The application reads an integer from the command line using args[0], converts it to an int, and uses it in a loop. When testing, the application throws a NumberFormatException when the user provides an alphabetic string. The developer needs to handle this exception gracefully by prompting the user to enter a valid number and retrying. However, the developer must avoid infinite loops. The current code uses a while loop with a flag. Which approach ensures the code handles the exception, provides feedback, and terminates if the user enters 'quit'? The environment is a standard Java SE 11 application. The developer wants a robust solution without using external libraries.

Accepted Answer

Use a try-catch inside a loop that continues if exception caught, and breaks if input equals 'quit' or conversion succeeds.. It uses a try-catch inside a while loop that continues if an exception is caught, allowing the user to re-enter input. The loop also checks if the input equals 'quit' to break out, ensuring the program can terminate gracefully. This approach provides feedback (prompting the user) and avoids infinite loops. Option B is incorrect because ignoring the exception would not provide feedback or allow retry. Option C is incorrect because it never checks for 'quit', risking an infinite loop. Option D is incorrect because calling System.exit(0) terminates the program abruptly, which is not graceful.

Answer

Catch the exception and ignore it; the loop will naturally terminate.

Answer

Use a do-while loop that never checks for 'quit'; the exception is caught outside the loop.

Answer

Catch the exception in the loop and call System.exit(0) immediately.

Question 4

A developer declares an integer variable inside a method but does not assign a value. What is the result of attempting to print the variable?

Accepted Answer

Compilation error. In Java, local variables (declared inside a method) must be initialized before use. The compiler performs definite assignment analysis and will reject code that attempts to read an uninitialized local variable, producing a compilation error. This rule ensures memory safety and prevents undefined behavior.

Answer

Prints null

Answer

Prints 0

Answer

Runtime exception

Question 5

Given two String objects s1 = "Hello" and s2 = "Hello", what does the expression (s1 == s2) return?

Accepted Answer

true. The `==` operator compares object references, not string content. Since both `s1` and `s2` are string literals, Java's string interning ensures they refer to the same `String` object in the string constant pool, so `s1 == s2` returns `true`.

Answer

Compilation error

Answer

false

Answer

It depends on the JVM

Question 6

A developer writes: int x = 5; int y = x++ + ++x; What is the value of y after execution?

Accepted Answer

12. The expression `int y = x++ + ++x;` involves post-increment and pre-increment operators. Initially, `x = 5`. In `x++`, the current value (5) is used, then `x` becomes 6. In `++x`, `x` is incremented to 7, then the new value (7) is used. So, `y = 5 + 7 = 12`. Option A is correct.

Answer

11

Answer

10

Answer

Compilation error

Question 7

A developer wants to iterate over an array of integers named 'numbers'. Which loop declaration will correctly access each element?

Accepted Answer

for (int num : numbers). It uses the enhanced for-each loop syntax, which is designed specifically for iterating over arrays and collections in Java. The colon (:) separates the element variable declaration from the array name, and each iteration assigns the next element to 'num' automatically, without needing an index or explicit bounds checking.

Answer

for (int num = 0; num < numbers.length; num++)

Answer

for (int num in numbers)

Answer

for (int num : numbers.length)

Question 8

A class has two methods: void setValue(int a) and void setValue(double a). Which call will result in a compilation error?

Accepted Answer

setValue(5); setValue(5.0); both in same class. Since the class contains both overloaded methods void setValue(int a) and void setValue(double a), all the provided calls are valid and compile without error. setValue(5) calls the int version, setValue(5.0) calls the double version, and setValue((int)5.0) explicitly casts to int, calling the int version. Option B consists of two separate valid calls within the same class, so no compilation error occurs. Therefore, none of the options result in a compilation error.

Answer

setValue(5.0);

Answer

setValue(5);

Answer

setValue((int)5.0);

Question 9

A parent class has a static method display() and an instance method show(). A child class attempts to override both. What is the outcome?

Accepted Answer

Only show() can be overridden. In Java, static methods belong to the class, not instances, so they cannot be overridden—they can only be hidden. Instance methods, however, can be overridden to provide polymorphic behavior. Therefore, only the instance method show() can be overridden; display() is hidden, not overridden.

Answer

Both will be overridden

Answer

Only display() can be overridden

Answer

Neither can be overridden

Question 10

Which THREE are valid Java identifiers?

Accepted Answer

my_value. Option B, my_value, is a valid Java identifier because it begins with a letter and contains only letters and the underscore character, which are legal identifier parts. Option C, $dollar, is valid because the dollar sign ($) is explicitly allowed as a Java identifier start character (and as a subsequent character). Option E, _myVar, is valid because the underscore (_) is a legal Java identifier start character, so an identifier may begin with it. Option A, 2ndValue, is invalid because a Java identifier cannot begin with a digit. Option D, my-var, is invalid because the hyphen (-) is not a legal character in a Java identifier; only letters, digits, $, and _ are permitted.

Answer

2ndValue

Answer

my-var

Question 11

Which THREE statements about the final keyword in Java are true?

Accepted Answer

A final variable cannot be reassigned after initialization.. Option A is correct because a final variable can be assigned only once, so any attempt to reassign it after initialization causes a compile-time error. Option B is correct because final applies to the reference itself, not the referenced object: the reference must keep pointing to the same object, although the object's internal state can still be mutated. Option E is correct because declaring a class final prevents any other class from extending it, which is why classes like java.lang.String are final. Option C is incorrect because a final method cannot be overridden in a subclass (it can only be inherited and used as-is). Option D is incorrect because a final parameter cannot be reassigned inside the method, so the method cannot modify the argument value.

Answer

A final method can be overridden in a subclass.

Answer

A final parameter in a method allows the method to modify the argument value.

Question 12

A junior developer wrote the following code to calculate the sum of an array:

int[] numbers = {1, 2, 3, 4, 5};
int sum = 0;
for (int i = 1; i <= numbers.length; i++) {
    sum += numbers[i];
}
System.out.println("Sum: " + sum);

The developer expects the output to be 15, but the program throws an exception. What is the root cause and the correct fix?

Accepted Answer

Change to: for (int i = 0; i < numbers.length; i++). The loop starts at i = 1 and runs while i <= numbers.length, so when i equals numbers.length (5), it accesses numbers[5], which is out of bounds for a 5-element array (valid indices 0–4), causing an ArrayIndexOutOfBoundsException. The correct fix is to start at i = 0 and use i < numbers.length to iterate all elements exactly once.

Answer

Change to enhanced for loop: for (int num : numbers) sum += num;

Answer

Change to: for (int i = 1; i < numbers.length; i += 2)

Answer

Change to: for (int i = 0; i <= numbers.length; i++)

Question 13

A team is developing a library management system. They have a base class 'Item' with a method 'getTitle()' that returns the title. They also have a subclass 'Book' that overrides 'getTitle()'. In some places, they have a method that accepts an Item reference but actually receives a Book object. They want to call a method specific to Book, such as 'getISBN()', that is not in Item. They attempt to cast the parameter: ((Book) item).getISBN(). However, occasionally the program crashes with ClassCastException. What is the best practice to avoid this exception?

Accepted Answer

Use if (item instanceof Book) before casting.. Using `instanceof` before casting ensures the object is actually an instance of `Book`, preventing a `ClassCastException` at runtime. This is the standard Java practice for safe downcasting when the object type is not guaranteed.

Answer

Create a separate method for Book objects.

Answer

Catch ClassCastException and handle it.

Answer

Change the parameter type to Object and then cast to Book.

Question 14

A mobile app backend uses Java streams to process user data. The code snippet filters users who are active and older than 18, then collects them into a list:

List<User> result = users.stream()
        .filter(u -> u.isActive())
        .filter(u -> u.getAge() > 18)
        .collect(Collectors.toList());

Performance metrics show that this stream operation is slow when the user list has millions of entries. The team wants to improve performance without changing the business logic. Which change would most likely improve performance?

Accepted Answer

Replace .stream() with .parallelStream().. Replacing `.stream()` with `.parallelStream()` enables parallel processing of the stream, which can leverage multiple CPU cores to process the filter operations concurrently. For large datasets (millions of entries), this can significantly reduce execution time by splitting the workload across threads, while preserving the same business logic and order of operations.

Answer

Use a HashSet instead of List for users collection.

Answer

Replace the stream with a traditional for loop.

Answer

Combine both filters into one using logical AND operator.

Question 15

Which TWO of the following are valid Java identifiers? (Choose two.)

Accepted Answer

$value. Option B, $value, is a valid Java identifier because Java allows the dollar sign ($) as an identifier start and part character, in addition to letters, the underscore, and Unicode currency symbols. Option C, _myVariable, is also valid because an underscore (_) is a legal Java identifier start character, and it is followed by valid identifier characters. Option A, my-var, is invalid because the hyphen (-) is not permitted in Java identifiers; it would be parsed as subtraction. Option D, 2ndPlace, is invalid because a Java identifier cannot begin with a digit. Option E, class, is invalid because class is a reserved Java keyword and cannot be used as an identifier.

Answer

my-var

Answer

2ndPlace

Answer

class