The process of erasure in Java generics replaces the type parameter with its bound or Object if no bound is specified. In this case, the type parameter is `<T>`, so the erasure of the method signature will replace `<T>` with `Object`. The correct erasure for the given code snippet is option A.
The provided Java code snippet defines a generic method `fun` with a type parameter `<T>`. During the erasure process by the compiler, the type parameter is replaced with its bound, or `Object` if no bound is specified. Therefore, the correct erasure for the method signature is `public static <T> void fun(Object[] t) { . . . }`.
This means that the method can accept an array of objects of any type. Options B, C, and D do not accurately reflect the correct erasure and are therefore incorrect. Understanding erasure is essential in Java generics, as it helps maintain compatibility with older code that doesn't use generics.
Therefore, option A is the correct answer.
The complete question is:
Consider the following code snippet:
Public static <T> void fun(T[] t) { . . . }
Erasure by the compiler of method fun will generate which result?
A) public static <T> void fun(Object[] t) { . . . }
B) public static <Object> void fun(Object t) { . . . }
C) public static void fun(Object[] t) { . . . }
D) public static void fun(Object t) { . . . }