Method oveloading - Object vs Object vararg

Question

See the code below:

// 1st method
private static void method(Object o){
    System.out.println(\"object method\");
}
// 2nd method
private static void method         


        

Answer 1

JLS 15.12.2 explains this exact scenario:

The first phase (§15.12.2.2) performs overload resolution without permitting boxing or unboxing conversion, or the use of variable arity method invocation. If no applicable method is found during this phase then processing continues to the second phase.

This guarantees that any calls that were valid in the Java programming language before Java SE 5.0 are not considered ambiguous as the result of the introduction of variable arity methods, implicit boxing and/or unboxing. However, the declaration of a variable arity method (§8.4.1) can change the method chosen for a given method method invocation expression, because a variable arity method is treated as a fixed arity method in the first phase. For example, declaring m(Object...) in a class which already declares m(Object) causes m(Object) to no longer be chosen for some invocation expressions (such as m(null)), as m(Object[]) is more specific.

Answer 2

Because Object... is essentially Object[] and since null is a valid Object[], it will match to the most specific one.

If you had 3 methods with first having Object parameter, the second having SubClass param and the last having SubSubClass param, the last one would be chosen.

Whereas if you add a method with a String parameter to your original code, you will get a compile time error, since there is no longer a single most specific match for null.