Initializing array on the fly in constructor (Java)

Question

 public class A{
      public A(int[] a){}
 }

 public class B extends A{
      public B(double[] b){
           super({b.length});  //ERROR
      }
 }

I want to be able to compile the code above. To clarify, I have class A and B that extends it. Class A does not have an empty parameter constructor. If I don't put a call to super in Class B's constructor on the first line, it will try to call super(), which doesn't exist. But, I want to call super(int[] a) instead. I want to do this by taking the length of a given double array and sending it as an array with length 1. It does not let me do this because apparently you can't declare an array like that, and if I were to declare it on a separate line it would call super() first and that won't work.

Is there any way to declare an int[] on the fly like that? Or are the only solution here to either make a constructor for A with no parameters or make my own function that returns an int[]?

(Don't ask why I want to send it as an array like that.)

corsiKa · Accepted Answer · 2011-09-29 00:42:46Z

11

If you insist on not asking why...

You could make the array, assign the first and only element and send it.

public class B extends A{
      public B(double[] b){
           int[] arr = new int[1];
           arr[0] = b.length;
           super(arr);  // broken, super must be first.
      }
}

This means you must have a one line solution. Luckily, Java provides an in-line way to make a series of elements into an array at compile time.

public class B extends A{
      public B(double[] b){
           super(new int[]{b.length});  // FIXED
      }
}

edited Sep 29, 2011 at 0:42

answered Sep 29, 2011 at 0:34

corsiKa

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4 Comments

user828835 Over a year ago

Thank you, I did not know you can declare and initialize arrays like that at the same time. That is just what I was looking for.

user828835 Over a year ago

HOWEVER, your first solution would still not work because you're missing a call to super() in the first line of the constructor, so it would be called automatically. But since I lack a constructor A(), this would cause an error. Thanks anyway, though.

user802421 Over a year ago

The first example won't compile. You need to call super/this constructor at the very first line. Second exp +1.

OscarRyz Over a year ago

You can't call super like that in your first code. The correct answer is the second one

irreputable · Accepted Answer · 2011-09-29 00:36:46Z

4

you can also

 public class A{
    public A(int... a){}
 }

 public class B extends A{
    public B(double[] b){
       super( b.length ); 
  }
 }

answered Sep 29, 2011 at 0:36

irreputable

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Comments

OscarRyz · Accepted Answer · 2011-09-29 00:38:14Z

4

Yeap, try:

 super(new int[]{b.length});  //ERROR NO MORE

edited Sep 29, 2011 at 0:38

answered Sep 29, 2011 at 0:33

OscarRyz

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1 Comment

user828835 Over a year ago

No, that would just make the length of the array equal to b.length. I want the length to be 1, and the value of the first (and only) entry to be b.length.

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