Why is the change of an array argument in a method is reflected in the calling method if Java is pass-by-value?

Asked Feb 16 '15 at 06:55

Active Feb 18 '15 at 04:20

Viewed 41 times

Why can a calling function see the modified value of an array after it's passed to a called method? Since an array is an object, I'd assume it behaves the same as an object, but it doesn't. Can anyone explains why?

public class test {
  public static void funk(int bird, int[] tiger) { 
    tiger[0] = tiger[0] * 2; 
    bird = tiger[0] + 5;
  } 
  public static void main(String[] args) {
    int bird = 10;
    int[] tiger = {7};
    Test.funk(bird, tiger);

    //bird: 10 --> Not changed
    System.out.println("bird:"+bird);           
    //tiger: [14] --> Changed 
    System.out.println("tiger:"+Arrays.toString(tiger));
  }
}

edited Feb 18 '15 at 04:20

asked Feb 16 '15 at 06:55

LED Fantom

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`int[]` is not a primitive type. – Sotirios Delimanolis Feb 16 '15 at 06:56
@SotiriosDelimanolis - The OP has already mentioned that he has gone through the post used to mark this question as *duplicate*. :) – TheLostMind Feb 16 '15 at 06:57
1

@TheLostMind That doesn't make it any less of a duplicate. [This](http://stackoverflow.com/a/18740042/438154) answer addresses arrays. – Sotirios Delimanolis Feb 16 '15 at 06:58
@SotiriosDelimanolis - You are right. He should go through that post *again* :) – TheLostMind Feb 16 '15 at 07:00
What makes you think it doesn't behave like an object? – Sotirios Delimanolis Feb 18 '15 at 06:54

Why is the change of an array argument in a method is reflected in the calling method if Java is pass-by-value?

0 Answers0