Set variable at compile time

[Craig Dillabaugh]

On Wednesday, 30 October 2013 at 20:23:58 UTC, Ali Çehreli wrote:
> On 10/30/2013 01:11 PM, Craig Dillabaugh wrote:
>
> > I am writing code that uses a structure containing an array of
> > points where the points may be of arbitrary dimension (though
> > generally small).  I would like to be able to pass the point
> > dimension to my structure as a template parameter.
>
> struct Point
> {}
>
> struct S(size_t N)
> {
>     Point[N] points;
> }
>
> alias S2D = S!2;
> alias S3D = S!3;
>
> void main()
> {}
>
> > One solution is to create instances of these structures for
> all
> > reasonable dimensions and then select the correct instance at
> run
> > time I suppose.
>
> Do you need a common interface, or all of the algoritms will be 
> templatized as well? In other words, what is the type that the 
> following function returns?
>
> ??? selectInstance(size_t N)
> {
>     // return S2D or S3D?
> }
>
> You can have an interface that the template implements:
>
> interface SInterface
> {
>     void foo();
> }
>
> class S(size_t N) : SInterface
> {
>     Point[N] points;
>
>     void foo() { /* implementation for N */ }
> }
>
> Now selectInstance() returns SInterface:
>
> SInterface selectInstance(size_t N)
> {
>     // ...
> }
>
> > However, I don't really want to set a limit on
> > the point dimension.
>
> All of the instances must be known at compile time. So, your 
> program is always limited with the number of actual instances 
> that the program is using.
>
> Ali

Ali thanks.

The algorithms will be templatized as well.  As you point out, I
need to know all instances at compile time, but want to 'get
around' this, thus my idea of compiling the specific instance I
need.  Adam's idea should work.

I like the interface idea, but just to make sure I understand the
purpose - you use the interface to provide a common interface so
that the instantiated objects can be used by algorithms which do
not have any knowledge of the particular dimension of a given
point.  Is that correct?

Craig