Re: Why does reverse also flips my other dynamic array?

[Marco Leise via Digitalmars-d-learn]

Thanks for the clarification.

Re: Why does reverse also flips my other dynamic array?

[Jonathan M Davis via Digitalmars-d-learn]

On Sunday, September 13, 2015 17:17:01 Ali Çehreli via Digitalmars-d-learn 
wrote:
> 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...

Yeah. Only dynamic arrays whose last element refers to the last element that
has been expanded into in a GC-allocated memory buffer can be appended to
without reallocating. If the memory block wasn't specifically allocated by
the GC for dynamic arrays, then there won't be any memory to expand into,
and even if it is a GC-allocated buffer for dynamic arrays (as is usually
the case), if another dynamic array has ever referred to any elements past
the end of the dynamic array you're looking at, then that dynamic array
can't be appended to without being reallocated, because it would potentially
stomp on another dynamic array. So, if you have multiple dynamic arrays
referring to the same memory, only one of them can expand into the unused
memory past their ends. Anything else would allow stomping.

The result of this is that if you're doing a lot of appending, you probably
don't want to be doing a lot of slicing unless the slices aren't being
appended to. It'll still work if they're appended to, but pretty quickly
it'll mean that most of the appending that you're doing is causing
reallocations, which could really harm performance.

- Jonathan M Davis

Re: 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

Re: Why does reverse also flips my other dynamic array?

[Ali Çehreli via Digitalmars-d-learn]

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);
}

Ali

Re: Why does reverse also flips my other dynamic array?

[Marco Leise via Digitalmars-d-learn]

Am Sat, 12 Sep 2015 10:55:50 +
schrieb "Namal" :

> > Why is also b flipped here? This doesn't happen if I use static 
> > arrays.
> 
> nvm. I need to .dup that.

Correct, static arrays are value types and copied on
assignment. Dynamic arrays on the other hand are generally
slices of memory on the garbage collected heap represented by
just a pointer to the first element and a length. When you
assign those you only get a second reference to the same data.

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". (This is so you can't accidentally
append different items to both a and b and have them overwrite
each other.) You can query the actual capacity with a.capacity.

Just felt like writing down some knowledge you might need
along the way. :)

-- 
Marco

Re: Why does reverse also flips my other dynamic array?

[Namal via Digitalmars-d-learn]

Why is also b flipped here? This doesn't happen if I use static 
arrays.


nvm. I need to .dup that.