Array of integers boxing

Question

Let's say I have an array of integers and attempt to do a call to CopyTo() onto a different array:

int[] a = {1, 2, 3};
int[] b = new int[3];

a.CopyTo(b, 0); // does b have a completely new set of { 1, 2, 3 }?

CopyTo() does a shallow copy, but I'm wondering, since int[] is a reference type, would this cause the elements within a to become boxed and thus no longer be value types?

Answer 1

Since you've created a new instance for array b

int[] b = new int[3];

the answer for your first question

does b have a completely new set of { 1, 2, 3 }?

is yes; b doesn't share the reference with a:

// false - a and b are different instances
// even if then can have same items they don't share the same reference
Console.Write(ReferenceEquals(a, b));

Since you've created b as an array of int - int[] there will be no boxing.; all the items will be copied by values. If you want to play with boxing make b keep items by reference: object[] b = new object[3];; if you want to see shallow copy make both arrays objects[]:

// boxed items
object[] a = new int[] { 1, 2, 3 };

// boxed items
object[] b = new object[3];

// references to boxed integers will be copied 
a.CopyTo(b, 0);

// True
Console.WriteLine(ReferenceEquals(a[0], b[0]));

// Let's demonstrate that a[0] and b[0] share the same reference:
// We change a[0] and we get b[0] changed as well:

// boxed values don't expose their inner data, let's use Reflection
a[0].GetType()
    .GetField("m_value", BindingFlags.Instance | BindingFlags.NonPublic)
    .SetValue(a[0], 456);

// 456
Console.WriteLine(b[0]);