T[new]

[Steven Schveighoffer]

On Sun, 09 Aug 2009 16:29:21 -0400, Walter Bright  
<newshound1 at digitalmars.com> wrote:

> D has a number of subtle problems (performance and semantic) that arise  
> when arrays are resized. The solution is to separate resizeable array  
> types from slices. Slices will retain the old:
>
>     T[] slice;
>
> syntax. Resizeable arrays will be declared as:
>
>     T[new] array;
>
> The new expression:
>
>     new T[10]
>
> will return a T[new].
>
> T[new] will implicitly convert to T[], but not the other way.
>
> slice.length will become read-only.
>
> Under the hood, a T[new] will be a single pointer to a library defined  
> type. This library defined type will likely contain three properties:
>
>      size_t length;
>      T* ptr;
>      size_t capacity;
>
> The usual array operations will work on T[new] as well as T[].
>
> Doing this change will:
>
> 1. fix many nasties at the edges of array semantics
>
> 2. make arrays implementable on .net
>
> 3. make clear in function signatures if the function can resize the  
> array or not

Yay!

Will capacity be settable?  That is, can I replace this pattern:

char[] buf = new buf[1024];
buf.length = 0;

with something like this?:

char[new] buf;
buf.capacity = 1024; // ensure capacity is *at least* 1024

----

I read elsewhere that

T[new] x;

doesn't allocate until length (or maybe capacity) is non-zero.

So T[new] is a library defined type.  What does the underlying type get  
called with when x.length = 5 is entered?  Is the underlying type:

1. an object?
2. a templated type?

----

What happens when you do:

x = null;

where x is a T[new] type?

Is it the same as setting x.length = 0, or x.capacity = 0?

What does x == null mean?  What about x is null?

Will ptr be settable, or only readable?

Will the compiler continue to optimize foreach calls to arrays?  How will  
foreach be implemented, as a range or opApply?  The former implies some  
automated method to return a range, because you wouldn't want to modify  
the referenced array during iteration.

This is exciting :)

-Steve