1Z0-819 Generics & Type Inference 3

Question 1

What is the var keyword available in Java 10 and Java 11?

Accepted Answer

A keyword for local variable type inference that lets the compiler determine the type

Answer

A keyword for declaring variable-length arguments in methods

Answer

A keyword for creating mutable variables that can be reassigned in lambda expressions

Answer

A keyword for declaring variables that can hold any numeric type

Question 2

Which of the following var declarations is INVALID in Java 11?

Accepted Answer

var x;

Answer

var list = new ArrayList<String>();

Answer

var number = 42;

Answer

var result = someMethod();

Question 3

Where can the var keyword NOT be used in Java 11?

Accepted Answer

As the declared type of an instance field in a class

Answer

As the type in an enhanced for-loop variable

Answer

As the type of a local variable inside a method body

Answer

As the type in a try-with-resources variable declaration

Question 4

Why can you NOT instantiate a generic type parameter with new T() in Java?

Accepted Answer

Due to type erasure, T is not available at runtime so the JVM cannot know which class to instantiate

Answer

Constructors are not permitted inside generic classes

Answer

The compiler prohibits the new keyword when T is a type parameter

Answer

Generic type parameters can only be instantiated using the reflection API

Question 5

Which of the following correctly relies on type inference for a generic method call?

Accepted Answer

List<String> list = Collections.emptyList();

Answer

List list = Collections.<>emptyList();

Answer

List list = Collections.emptyList();

Answer

List list = Collections.emptyList();

Question 6

At runtime, how can you check if an object is specifically a List<String>?

Accepted Answer

You cannot check for the parameterized type at runtime due to type erasure

Answer

Use: obj instanceof List<String>

Answer

Use: obj.getClass() == List<String>.class

Answer

Use: obj instanceof List && ((List<?>) obj).stream().allMatch(e -> e instanceof String)

Question 7

What is the inferred type of the variable 'map' in: var map = new HashMap>();?

Accepted Answer

HashMap<String, List<Integer>>

Answer

Map<String, List<Integer>>

Answer

HashMap<String, ArrayList<Integer>>

Answer

Object