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