Why does reverse also flips my other dynamic array?

[Ali Çehreli via Digitalmars-d-learn]

On 09/13/2015 08:21 AM, Jonathan M Davis via Digitalmars-d-learn wrote:
 > 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

Thanks for clarifying. My point was about capacity not being lost; 
rather, being used (owned) by the _first_ slice that gets extended. If 
we append to 'a' first, then 'a' extends in place and 'b' gets 
relocated. If we append to 'b' first, then vice versa...

Ali