dotnetasit
Profile Answers by dotnetasit

Feb 21st, 2008

I felt the need of this posting because I have seen postings (not only in this forum but also in many others) saying that "CopyTo() makes a deep copy and Clone() makes a shallow copy." This is absolutely wrong.

Both CopyTo() and Clone() make shallow copy. Clone() method makes a clone of the original array. It returns an exact length array.

On the other hand, CopyTo() copies the elements from the original array to the destination array starting at the specified destination array index. Note that, this adds elements to an already existing array.

The following code will contradict the postings saying that CopyTo() makes a deep copy:

public class Test
{
    public string s;
}

// Write Main() method and within it call test()

private void test()
{
    Test[] array = new Test[1];
    array[0] = new Test();
    array[0].s = "ORIGINAL";

    Test[] copy = new Test[1];
    array.CopyTo(copy, 0);

    // Next line displays "ORIGINAL"
    MessageBox.Show("array[0].s = " + array[0].s);
    copy[0].s = "CHANGED";

    // Next line displays "CHANGED", showing that
    // changing the copy also changes the original.
    MessageBox.Show("array[0].s = " + array[0].s);
}

Let me explain it a bit. If the elements of the array are of reference types, then the copy (both for Clone() and CopyTo()) will be made upto the first(top) level. But the lower level doesn't get copied. If we need copy of lower level also, we have to do it explicitly. That's why after Cloning or Copying of reference type elements, each element in the Cloned or Copied array refers to the same memory location as referred by the corresponding element in the original array. This clearly indicates that no separate instance is created for lower level. And if it were so then changing the value of any element in the Copied or Cloned array would not have effect in the corresponding element of the original array.

I think that my explanation is exhaustive but I found no other way to make it understandable. Hope this will help everyone.