1Z0-819 Generics & Type Inference

Question 1

What does type erasure mean in the context of Java generics?

Accepted Answer

Generic type parameters are converted to their bounds (or Object) in compiled bytecode

Answer

Type erasure means the compiler removes generic type information during compilation, replacing type parameters with their bounds or Object in the bytecode to maintain backward compatibility.

Question 2

Which of the following correctly declares a generic method in Java?

Accepted Answer

public <T> T getValue() { return null; }

Answer

A generic method must declare its type parameter(s) in angle brackets before the return type, as in public <T> T getValue().

Question 3

What does the diamond operator <> accomplish when used in Java?

Accepted Answer

Instructs the compiler to infer the type parameter from the surrounding context

Answer

The diamond operator <> tells the compiler to infer the generic type argument from the left-hand side or context, reducing verbosity in object creation expressions.

Question 4

Which statement about using raw types in Java is TRUE?

Accepted Answer

Raw types bypass generic type checking and can lead to ClassCastException at runtime

Answer

Raw types bypass the compiler's generic type safety checks, potentially leading to ClassCastException at runtime when incompatible types are inserted or retrieved.

Question 5

Given: List strList = new ArrayList<>(); Which assignment is valid?

Accepted Answer

List<? extends String> extList = strList;

Answer

List accepts a List because it is an upper-bounded wildcard; generic types are invariant, so all Object-based assignments are compile errors.

(1Z0-819) Oracle Java SE 11 Developer Practice Test

(1Z0-819) Oracle Java SE 11 Developer Practice Test

1Z0-819 Generics & Type Inference