std.array.put doesn't put

[Paul D. Anderson]

I've entered this as a Phobos bug, but wanted to be sure I was understanding 
this properly (i.e., this is a bug, right?):

>From the description of the put primitive in std.range:

"r.put(e) puts e in the range (in a range-dependent manner) and advances to the 
popFront position in the range. Successive calls to r.put add elements to the 
range. put may throw to signal failure."

>From the example of std.array for the put function:

void main()
{
    int[] a = [ 1, 2, 3 ];
    int[] b = a;
    a.put(5);
    assert(a == [ 2, 3 ]);
    assert(b == [ 5, 2, 3 ]);
}

So, "putting" 5 into the array a removes the first element in a, and changes 
the value of the first element of b. I would expect the first assert in the 
code above to read:

    assert(a == [ 5, 1, 2, 3 ]);

The implementation of std.array.put doesn't make sense: 

void put(T, E)(ref T[] a, E e) { assert(a.length); a[0] = e; a = a[1 .. $]; }

It modifies a[0] and then replaces the array with the tail of the array,
omitting the first element.

It's possible there is some arcane meaning to the word "put" that I'm not aware 
of, but if it means "insert an element at the front of the range" then 
std.array.put is wrongly implemented.

Paul