• Dominikus Dittes Scherkl (20/20) Nov 03 2014 If I have a struct with numeric template parameter, how can I
• MrSmith (4/24) Nov 03 2014 You cannot assign to it, because it is only avaliable during
• Philippe Sigaud via Digitalmars-d-learn (19/28) Nov 03 2014 base is part of the type. polynomial is just a 'recipe' for a type,
• Dominikus Dittes Scherkl (6/31) Nov 05 2014 Yes, that's what I intend.

"Dominikus Dittes Scherkl" writes:

If I have a struct with numeric template parameter, how can I 
access it within member functions? Like normal member variables? 
And how about the constructor?

struct polynomial(uint base)
{
private:
    uint[] N;
public:
    this(uint x) { base = x; }
    ...
    void add(Polynomial!base P)
    {
       if(N.length < P.N.length) N.length = P.N.length;
       foreach(i; 0..P.N.length)
       {
          N[i] = (N[i]+P.N[i]) % base;
       }
    }
}

This doesn't work for me :-/

"MrSmith" <mrsmith33 yandex.ru> writes:

On Monday, 3 November 2014 at 14:27:47 UTC, Dominikus Dittes 
Scherkl wrote:
 If I have a struct with numeric template parameter, how can I 
 access it within member functions? Like normal member 
 variables? And how about the constructor?

 struct polynomial(uint base)
 {
 private:
    uint[] N;
 public:
    this(uint x) { base = x; }
    ...
    void add(Polynomial!base P)
    {
       if(N.length < P.N.length) N.length = P.N.length;
       foreach(i; 0..P.N.length)
       {
          N[i] = (N[i]+P.N[i]) % base;
       }
    }
 }

 This doesn't work for me :-/

You cannot assign to it, because it is only avaliable during 
compilation. Think of it as an immediate value, not variable.

Philippe Sigaud via Digitalmars-d-learn writes:

On Mon, Nov 3, 2014 at 3:27 PM, Dominikus Dittes Scherkl via
Digitalmars-d-learn <digitalmars-d-learn puremagic.com> wrote:
 If I have a struct with numeric template parameter, how can I access it
 within member functions? Like normal member variables? And how about the
 constructor?

 struct polynomial(uint base)
 {
 private:
    uint[] N;
 public:
    this(uint x) { base = x; }

base is part of the type. polynomial is just a 'recipe' for a type,
the real struct would be Polynomial!(0), Polynomial!(1), etc. Note
that Polynomial!0, Polynomial!1, ... are all different types.

Being part of the type means it's defined only at compile-time, you
cannot use a runtime value (like 'x') to initialize it.

Note that with your current code, `base' is not visible outside
Polynomial. You can alias it to a field to make it visible:

struct Polynomial(uint base)
{
    alias b = base; // b is visible outside (but set at compile-time !)
...
}

You can create one like this:

Polynomial!2 poly;
poly.N = [0,1,0,0,1,1];

assert(poly.b == 2);

Of course, you cannot change b: `poly.b = 3;' is forbidden.

"Dominikus Dittes Scherkl" writes:

On Monday, 3 November 2014 at 21:17:09 UTC, Philippe Sigaud via
Digitalmars-d-learn wrote:
 struct polynomial(uint base)
 {
 private:
    uint[] N;
 public:
    this(uint x) { base = x; }

 base is part of the type. polynomial is just a 'recipe' for a 
 type,
 the real struct would be Polynomial!(0), Polynomial!(1), etc. 
 Note
 that Polynomial!0, Polynomial!1, ... are all different types.

Yes, that's what I intend.
 Being part of the type means it's defined only at compile-time, 
 you
 cannot use a runtime value (like 'x') to initialize it.

 Note that with your current code, `base' is not visible outside
 Polynomial. You can alias it to a field to make it visible:

 struct Polynomial(uint base)
 {
     alias b = base; // b is visible outside (but set at

Ah, ok. Thank you!

 compile-time !)
 ...
 }

 You can create one like this:

 Polynomial!2 poly;
 poly.N = [0,1,0,0,1,1];

Ok, now I remember, struct doesn't need an explicit constructor.
(in this case)