Varargs method modifies caller's array instead of its own copy?

Question

I have this simple varargs method which divides each item in the list: ``` import java.util.*; class A { static long f(long... xs) { Arrays.sort(xs); long y = 100000000; for (int i = xs.length - 1; i >= 0; i--) y /= xs[i]; return y; } static { System.out.println(f(5,2,6,3,9,3,13,4,5)); long[] xs = new long[]{5,2,6,3,9,3,13,4,5}; System.out.println(Arrays.toString(xs)); System.out.println(f(xs)); System.out.println(Arrays.toString(xs)); } } ``` I'd expect it to pass a copy of the array, but apparently it's somehow modifying the array I pass in, instead of its own local copy: ``` $ javac A.java && java A 79 [5, 2, 6, 3, 9, 3, 13, 4, 5] 79 [2, 3, 3, 4, 5, 5, 6, 9, 13] ``` So I wrote this simple test program: ``` class B { static void f(Object... os) { System.out.println(os); } static { Object os = new Object[]{1,2,3}; System.out.println(os); f(os); } } ``` And it does what I expect, it clones the object array before passing it into `f` (hence different object identifiers): ``` $ javac B.…

Accepted Answer

It looks like you've tricked yourself here:

```
Object os = new Object[]{1,2,3};
System.out.println(os);
f(os);
```

Since `os` is typed as `Object`, it gets interpreted as the first element of the varargs array. What gets passed into the method is actually a new `Object[]` whose single element is your `Object[]`.

If you do the following, it will print the same instance:

```
Object[] os = new Object[]{1,2,3};
System.out.println(os);
f(os);
```

The `f` method will need make a defensive copy of the array itself in order to guarantee that an array passed in by the caller isn't modified. As arshajii points out, varargs are foremost array parameters, with the "bonus" behavior of creating a new array when given an argument list.

Anyway you can use `Arrays.copyOf` to make the copy, which delegates to (the less type-safe) `System.arraycopy`.

Varargs method modifies caller's array instead of its own copy?

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