To answer this question, we need to understand method overloading and method invocation in Java.
In the given code, there are two methods in the Alien
class with the same name but different parameters:
1) String invade(short ships)
2) String invade(short... ships)
The first method takes a single short
parameter, while the second method takes a variable number of short
parameters.
In the Defender
class, the main
method creates a new instance of the Alien
class and invokes the invade
method with an argument of 7
.
Since 7
is an integer literal, it is implicitly converted to a short
before being passed as an argument to the invade
method.
Now, the Java compiler tries to find the most specific method to invoke based on the argument passed. In this case, both methods can accept a short
argument, so the method with a single short
parameter is a more specific match.
However, when there are multiple methods with the same name and same specificity of match, the Java compiler gives priority to methods without varargs. In this case, the method with varargs is considered less specific.
Hence, the compiler fails because it cannot determine which invade
method to invoke, resulting in a compilation error.
Therefore, the correct answer is option C) Compilation fails.