Java: extending an object to another

Question

For example I have this:

class A{
    private int mine = 0; //Some field...
    public A(int a){mine+=a; /*Some operation1...*/}
}


class B extends A{
    private int mine = 0; //Some field...
    public B(int a){mine-=a; /*Some operation2...*/}
}

and I get:

error: constructor A in class A cannot be applied to given types;
    public B(int a){}
    required: int
    found: no arguments
    reason: actual and formal argument lists differ in length
    1 errors

I don't understand the error? What is telling me to do?

Though, the code works if Constructor of "A" has no arguments.
But I need to do the Operation1 (aka mine+=a;), so I need A's arguments, but then I fail.

I'm closed in this magic circle. What shall I do?

Why mine+=a; and not just mine = a;?

Ash Burlaczenko
– Ash Burlaczenko

2012-05-06 22:12:45 +00:00
Commented May 6, 2012 at 22:12 — Ash Burlaczenko
– Ash Burlaczenko, Commented May 6, 2012 at 22:12

JB Nizet · Accepted Answer · 2012-05-06 22:13:14Z

9

The first instruction of every constructor is always to invoke one of its superclass constructor. If you don't do it explicitely, the compiler inserts this instraction for you. The constructor

 public B(int a) {
     mine-=a; 
     /*Some operation2...*/
 }

is thus equivalent to

public B(int a) {
    super(); // invoke the no-arg super constructor
    mine-=a; 
    /*Some operation2...*/
}

Since A doesn't have a no-arg constructor, the compilation fails. In this case, you must explicitely invoke one of the super constructors:

public B(int a) {
    super(a);
    mine-=a; 
    /*Some operation2...*/
}

answered May 6, 2012 at 22:13

JB Nizet

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Comments

Michael Borgwardt · Accepted Answer · 2012-05-06 22:12:48Z

7

Every constructor must call a superclass constructor as the first thing it does. If you do not do it explicitly via super(), the compiler will implicitly call the no-arts constructor in the superclass. But in your case, the superclass does not have a no-args constructor, at which point the compiler gives up.

By the way, it's generally a very bad idea to re-declare a field in a subclass, because it will hide the superclass field, but there will still be two fields.

answered May 6, 2012 at 22:12

Michael Borgwardt

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Comments

peter.murray.rust · Accepted Answer · 2012-05-06 22:13:21Z

3

I think you need to call class A's constructor from B constructor, like this: super(a);

edited May 6, 2012 at 22:13

peter.murray.rust

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answered May 6, 2012 at 22:12

Martin

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Comments

user unknown · Accepted Answer · 2012-05-06 22:13:39Z

If B extends A, the first operation of the constructor of B is, to call the constructor of A.

If there is no constructor explicitly defined in A, the standard constructor is called. If you don't call the ctor explicitly, it is called implicitly.

But if there is an explicit ctor, which takes arguments, it isn't called automatically - how should it - with which arguments? You have to call it yourself.

But your class hierarchie is somewhat misterious. If you want to access mine in the subclass, make it protected or leave it in default state - don't make it private.

If you make it private, a new introduced attribute with the same name hides the parent attribute, but this is errorprone and irritating.

In most cases, access, maybe via a method (getter/setter), would be better, or opening the visibility.

Or a different name for the thing in the subclass.

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