Static arrays and std.algorithm.sort

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I believe the problem is that std.algorithm.sort requires a
range, and fixed-size arrays are not ranges in D. I believe it's
because the most basic range functionality is the ability to do
this:

int[] arr = [0, 1, 2, 3];
while (arr.length > 0)
{
      arr = arr[1..arr.length];
}

In other words, shrink the range size, which is obviously not
possible with a fixed-size array, since its size is... fixed. The
reason your example code works:

void main ()
{
	int[5] a = [9, 5, 1, 7, 3];
	//sort (a); //Fails
          //Note that `a[0..a.length]` is equivalent to the shorter
syntax `a[]`
	//sort (a[0..a.length]); //Works
	writeln (a);
}

Is because while int[5] is not a range, int[] *is* a range, as I
demonstrated above. Although int[5] is not a range, you can
obtain a range "view" of it with the square brackets syntax. This
is referred to as slicing. The following are all valid slices of
an int[5]:

int[5] arr = [0, 1, 2, 3];
auto slice1 = arr[0..$]; //$ means arr.length in this context.
                           //slice1 == [0, 1, 2, 3]

auto slice2 = arr[1..$]; //slice2 == [1, 2, 3]

auto slice3 = arr[1..3]; //slice3 == [1, 2]

auto slice4 = arr[0..0]; //slice4 == [] empty slice, length == 0
but ptr != null