Java: Find out whether function was called with varargs or array

Question

Is there a way to find out whether a Java function (or a constructor) that takes varargs was actually called with varargs or with an array?

Say I have the following:

Answer 1

No, you can't. It's meant to be entirely transparent - this code:

new MyCompositeObjects(a, b);

is exactly equivalent to

new MyCompositeObjects(new MyObjects[] { a, b });

If you can trust your callers to do the right thing, you could always create two static methods and make the constructor private:

public static MyCompositeObjects createWithCopy(MyObjects[] values) {
    return new MyCompositeObjects(Arrays.copyOf(values, values.length));
}

public static MyCompositeObjects createWithoutCopy(MyObjects... values) {
    return new MyCompositeObjects(values);
}

private MyCompositeObjects(MyObjects[] values) {
    this.objects = values;
}

Note how the "with copy" version doesn't use varargs, which should help users to use the right version.

Answer 2

The only way to know is to parse the code. This is because varargs is basicaly a compile time feature and doesn't change how the program runs.

If in doubt, I would always copy the array. Unless you know this will be a performance issue.

BTW: You can do the following.

MyCompositeObjects(MyObjects o1, MyObjects... objects) {

MyCompositeObjects(MyObjects[] objects) {

However, this likely to do the opposite of what you want.

Another option is to use a static factory.

private MyCompositeObjects(MyObjects[] objects) {
   this.objects = objects;
}

MyCompositeObjects create(MyObjects... objects) {
    return new MyCompositeObjects(objects.close());
}

MyCompositeObjects createNotCopied(MyObjects... objects) {
    return new MyCompositeObjects(objects, false);
}

Use the more cumbersome method name for the less safe version. This means if a method is chosen without much thought, the safe version is more likely to be used.