How can one determine the recursion depth for templated types?

Example code:

import std.stdio;
class Set(T){
  override string toString(){
    return "Set";
  }
}
void main(){
  auto s0= new Set!uint;
  writeln( s0); // writes Set

  auto s1= new Set!(Set!uint);
  writeln( s1); // should write SetSet
}

On Friday, 22 May 2015 at 21:13:50 UTC, Manfred Nowak wrote:
> How can one determine the recursion depth for templated types?
>
> Example code:
>
> import std.stdio;
> class Set(T){
>   override string toString(){
>     return "Set";
>   }
> }
> void main(){
>   auto s0= new Set!uint;
>   writeln( s0); // writes Set
>
>   auto s1= new Set!(Set!uint);
>   writeln( s1); // should write SetSet
> }

Why should the last line write SetSet? (It doesn't.) toString isn't making any use of the template argument T, so no recursion occurs.

Matt Kline wrote:

> isn't making any use of the template argument T

Correct. I do not know how to use `T' to determine the recursion depth of the template---and I want no further parameter.

-manfred

On 05/23/2015 12:12 AM, Manfred Nowak wrote:
> Matt Kline wrote:
>
>> isn't making any use of the template argument T
>
> Correct. I do not know how to use `T' to determine the recursion depth of
> the template---and I want no further parameter.
>
> -manfred
>

import std.stdio, std.range, std.algorithm;

template getDepth(T){
    static if(is(T==Set!S,S)) enum getDepth=1+getDepth!S;
    else enum getDepth=0;
}
class Set(T){
    override string toString(){
        enum r="Set".repeat.take(getDepth!Set).join;
        return r;
    }
}
void main(){
    auto s0=new Set!uint;
    writeln(s0); // writes Set
    auto s1=new Set!(Set!uint);
    writeln(s1); // writes SetSet
}

Timon Gehr wrote:

> template getDepth(T){
>      static if(is(T==Set!S,S)) enum getDepth=1+getDepth!S;
>      else enum getDepth=0;
> }

Thx. Seems that I have to relearn a lot.

-manfred