how does range.put work

[Daniel Keep]

Jos van Uden wrote:
> Oliver wrote:
>>> The source code for the standard library comes with the compiler.
>>> If you look in std\array.d, you find this around line 279 (reflowed for
>>> readability):
>>>> void put(T, E)(ref T[] a, E e) {
>>>>     assert(a.length);
>>>>     a[0] = e; a = a[1 .. $];
>>>> }
> 
> Would anybody care to explain what this is used for? I find
> the example in array.d rather unhelpful.
> 
> Example:
> ----
> void main()
> {
>     int[] a = [ 1, 2, 3 ];
>     int[] b = a;
>     a.put(5);
>     assert(a == [ 2, 3 ]);
>     assert(b == [ 5, 2, 3 ]);
> }
> 
> You're putting an element in a, but then the first element is moved out
> of a and the new one shows up in b? Weird. I guess I don't understand
> what a range is.
> 
> Jos

No; read the code.  Before the put, a and b are pointing to the same
span of memory.  a.put(5) puts the value 5 into the front (first
element) of the array, then advances the array.

However, put can't "see" b, so it doesn't get updated along with a.  The
end result is that b = [5,2,3] and a = b[1..3] = [2,3].

Why do it like this?  Here's an example:

void putNumbers(Range)(Range r)
{
    int i = 0;
    while( !r.empty )
    {
        r.put(i);
        ++i;
    }
}

void main()
{
    int[10] ten_numbers;
    putNumbers(ten_numbers);
    assert( ten_numbers = [0,1,2,3,4,5,6,7,8,9] );
}

Note that putNumbers will work with any type that supports the range
API, not just arrays.