On 12/09/2011 09:32 PM, Ali Çehreli wrote:
On 12/08/2011 12:52 PM, Timon Gehr wrote:
> On 12/08/2011 09:50 PM, RenatoL wrote:
>> snippet 1)
>> auto arr1 = [1,2,3];
>> auto arr2 = arr1;
>> arr1[1] = 22;
>> arr2[2] = 33;
>> foreach (i; 0..arr1.length) write(arr1[i], " ");
>> writeln();
>> foreach (i; 0..arr2.length) write(arr2[i], " ");
>>
>> output:
>> 1 22 33
>> 1 22 33
>> OK
>>
>> snippet 2)
>>
>> int[3] arr1 = [1,2,3];
>> int[3] arr2 = arr1;
>> arr1[1] = 22;
>> arr2[2] = 33;
>> foreach (i; 0..arr1.length) write(arr1[i], " ");
>> writeln();
>> foreach (i; 0..arr2.length) write(arr2[i], " ");
>>
>> output:
>>
>> 1 22 3
>> 1 2 33
>>
>> that's unclear to me... i "agree" with the behaviour of the
>> dynamic array... but if we have a static array we have a deep copy?
>
> Both copies are 'shallow', but static arrays are value types.
'shallow' would be misleading for a fixed-length (static) array because
there is nothing else but the elements for fixed-length arrays:
It is not misleading since the array might be an array of references.
(and if it does not contain references, shallow and deep are the same
thing anyway)
void main()
{
int[3] a;
assert(cast(void*)&a == cast(void*)&a[0]);
}
Fixed-length array storage is similar to C arrays. These are different:
- they don't decay to a 'pointer to first element' when passed to
functions (being value types, the whole array is copied)
- a.length is a convenience, equivalent to a.sizeof / a[0].sizeof
So it is impossible to do anything shallow with them unless we
explicitly maintain a pointer to a fixed-length array ourselves.
Ali
You can always slice it, of course
int[3] a;
int[] b = a[]; // b now is a dynamic array that aliases a's contents
