Question

Consider the following code snippet:

public class Box
{
private E data;
public Box() { . . . }
public void insert(E value) { . . . }
}
What will result from executing the following code?

Box box = new Box();
box.insert("blue Box");
A. compile-time error
B. run-time error
C. correct generic assignment
D. compile-time warning

Answer 1

The code using a raw generic type Box and calling insert will cause a compile-time warning, suggesting use of generics to ensure type safety.

Step-by-step explanation:

The student is working with Java generics and is testing the behavior of a generic Box class with the method insert. Given that the Box class is declared without specifying its generic type during instantiation, the expected behavior when calling box.insert("blue Box") would be a compile-time warning. This is because the code snippet uses a raw type of the generic Box class and does not specify the type parameter (e.g., Box<String>). The code will still compile, but the compiler can generate a warning indicating that the use of raw types is unsafe and should be avoided.