How does Java choose which constructor to use?

Question

I am not able to understand the output of the following program.

public class Confusing {

    private Confusing(Object o) {
        System.out.println("Object");
    }

    private Confusing(double[] dArray) {
        System.out.println("double array");
    }

    public static void main(String[] args) {
        new Confusing(null);
    }
}

The correct output is "double array". WHy was this constructor chosen as more specific than the other when both can accept null?

possible duplicate of [Which overload will get selected for null in Java?](http://stackoverflow.com/questions/1545501/which-overload-will-get-selected-for-null-in-java) — João Silva, Aug 21 '12 at 14:41

score 18 · Answer 1 · answered Aug 21 '12 at 14:41

18

Even though both constructors can accept null, double[] inherits from java.lang.Object, and is therefore more specific.

answered Aug 21 '12 at 14:41

Sergey Kalinichenko

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@downvoter would you care to elaborate on what you think is wrong with the answer? – Sergey Kalinichenko Aug 21 '12 at 14:53
1

Maybe because of [this](http://en.wikipedia.org/wiki/Resentment) (disclaimer: wasn't me) – David Aug 21 '12 at 14:59
I read this in Liskov's book "A method m1 is more specific than another method m2 if any legal call of m1 would also be a legal call of m2 if more conversions were done". This mean if there is a method comp(int) and comp(long), passing an int param will go to comp(int) (which is m1 above) as it could also go to comp(long) (which is m2 above) with conversion. But a vice versa does not hold as a comp cannot be passed to m1. However, how can you apply this logic to a null param in my original question - it can be applied to both double[] and Object without conversion?? – user1614482 Aug 21 '12 at 15:21
1

@user1614482 `int` and `long` are primitives, with conversion rules defined for them; unlike `Object` and `double[]`, neither one is a subclass of the other. When overloads use a single parameter related as base-subclass, the subclass wins when `null` is passed. – Sergey Kalinichenko Aug 21 '12 at 15:25

score 1 · Answer 2 · answered Aug 21 '12 at 14:44

The challenge of compiling dynamically typed languages is how to implement a runtime system that can choose the most appropriate implementation of a method or function — after the program has been compiled. Treating all variables as objects of Object type would not work efficiently.

Hence, choosing the specific one over Object.

How does Java choose which constructor to use?

2 Answers2

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