1Z0-811 Java Basics and Syntax Practice Test 6 — 15 Questions

Question 1

Given the loop: for (int i=0; i<5; i++) { if (i==2) continue; System.out.print(i); } What is the output?

Accepted Answer

0134. The loop iterates i from 0 to 4. When i equals 2, the 'continue' statement skips the rest of that iteration, so 'System.out.print(i)' is not executed for i=2. Therefore, the output is 0, 1, 3, 4 concatenated as '0134'. Options A and C both display '0134' and are therefore correct. Option B shows '0124' (missing 3) and Option D shows '01234' (includes 2), both incorrect.

Answer

0124

Answer

01234

Question 2

Which data type should be used to store a single character like 'A'?

Accepted Answer

char. In Java, the `char` data type is a single 16-bit Unicode character, capable of storing any character from the Unicode standard, including 'A'. It is the primitive type specifically designed for a single character, whereas `String` is a reference type for sequences of characters.

Answer

String

Answer

byte

Answer

int

Question 3

What is the result of the following code? String s1 = "Hello"; String s2 = " World"; String s3 = s1 + s2; System.out.println(s3);

Accepted Answer

Hello World. The + operator performs string concatenation in Java. s1 + s2 combines "Hello" and " World" to produce "Hello World", which is then printed. The space is part of s2, so the output includes it.

Answer

HelloWorld

Answer

Hello World

Answer

Compilation error

Question 4

Given: int[] arr = {10,20,30}; System.out.println(arr[3]); What is the result?

Accepted Answer

ArrayIndexOutOfBoundsException. Arrays in Java are zero-indexed, so a valid index for an array of length 3 is 0, 1, or 2. Accessing index 3 is out of bounds, causing an ArrayIndexOutOfBoundsException at runtime.

Answer

30

Answer

0

Answer

NullPointerException

Question 5

Which method overloading is valid?

Accepted Answer

void print(int a) and void print(String s) {}. Method overloading in Java requires methods to have the same name but different parameter lists (different number, type, or order of parameters). Here, `void print(int a)` and `void print(String s)` have different parameter types (`int` vs `String`), which satisfies the overloading rule. The return type or access modifiers do not affect overloading.

Answer

void print(int a) and void print(int b) {}

Answer

void print(int a) and int print(int a) {}

Answer

void print(int a) and private void print(int a) {}

Question 6

What is the scope of a variable declared inside a for loop?

Accepted Answer

Only inside the loop block. In Java, a variable declared inside a for loop (including the initialization block or the loop body) has block scope, meaning it is only accessible within the loop block itself. This is defined by the Java Language Specification (JLS §6.10), which states that the scope of a local variable declaration is the rest of the block in which the declaration appears. Once the loop completes, the variable goes out of scope and cannot be referenced.

Answer

Inside the package

Answer

Inside the class

Answer

Inside the entire method

Question 7

Which statement about try-catch is true?

Accepted Answer

A catch block can only handle one type of exception.. In Java, a catch block is defined with a single exception type parameter, meaning it can only handle one specific type of exception (or its subclasses if the parameter is a superclass). This is enforced by the language syntax: `catch (ExceptionType e)` allows only one exception type per catch clause.

Answer

A finally block is always required.

Answer

A try block without a catch or finally is legal.

Answer

A try block can have multiple catch blocks for the same exception type.

Question 8

Which TWO keywords are used to control access to class members? (Choose two.)

Accepted Answer

private. The `private` keyword restricts access to the class member so that it can only be accessed within the same class. Option C is correct because the `public` keyword allows access to the class member from any other class in any package. These are two of the four access modifiers in Java (private, default, protected, public) that directly control visibility and access to class members.

Answer

static

Answer

abstract

Answer

final

Question 9

Which THREE are primitive data types in Java? (Choose three.)

Accepted Answer

double. `double` is one of the eight primitive data types in Java, used to represent double-precision 64-bit IEEE 754 floating-point numbers. Primitives are predefined by the language and stored directly in stack memory, not as objects.

Answer

Integer

Answer

String

Question 10

The code does not compile. What is the error?

Accepted Answer

Assignment instead of comparison in if condition. The code uses a single equals sign `=` (assignment) inside the `if` condition instead of `==` (comparison). In Java, `if (x = 5)` assigns the value 5 to `x` and then evaluates the assignment expression to the assigned value (5), which is an `int`, not a `boolean`. The `if` statement requires a `boolean` expression, so this causes a compilation error.

Answer

Missing semicolon after if condition

Answer

Variable x is not initialized

Answer

System.out.println syntax error

Question 11

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 12

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 13

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 14

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 15

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