Why does reverse also flips my other dynamic array?

[Jonathan M Davis via Digitalmars-d-learn]

On Saturday, September 12, 2015 14:59:23 Ali Çehreli via Digitalmars-d-learn wrote:
> On 09/12/2015 02:29 PM, Marco Leise wrote:
>
>  > Note that often the original dynamic array has additional
>  > capacity beyond its length. This can be used to ~= additional
>  > items without causing a reallocation, but is lost when you
>  > do the assignment "b = a".
>
> Actually, the capacity is still there, useful to the runtime.
> Interestingly, the first dynamic array that is appended the new element
> becomes the owner of that capacity. The capacity of the other dynamic
> array becomes 0.
>
> import std.stdio;
>
> void main(){
>
>    int [] a = [1,2,3,4,5];
>    int [] b = a;
>
>    writeln(a.ptr, " ", b.ptr);
>    writeln(a.capacity, " ", b.capacity);
>    a ~= 42;    // <-- change to b, now b owns the capacity
>    writeln(a.ptr, " ", b.ptr);
>    writeln(a.capacity, " ", b.capacity);
> }

It's not really the case that the capacity is owned. There simply is no
available memory passed the end of b, because what would be the next element
in b if it were appended to is now the last element of a. It's the same boat
you're in if you simply did

int[] a = [1, 2, 3, 4, 5];
int[] b = a[0 .. $ - 1];

In both cases, there's no room for b to grow into, because a is using the
space immediately after b.

- Jonathan M Davis