I want to express in D the known Haskell qsort 3 lines code (it is not a quick sort, but an example of how functional programming is expressive).

This is the "javascript" version I use as reference:

    const sorted = ( [pivot, …others] ) => pivot===void 0 ? [ ] : [
      … sorted( others.filter( n=> n<pivot ) ),
      pivot,
      … sorted( others.filter( n=> n>=pivot ) )
    ];


This is a possible D implementation:

void main()
{
    import std.stdio, std.algorithm, std.range;
	
    int[] delegate(int[]) qs;
	
    qs = (int[] items) => items.length==0 ? items :
      chain(
        qs(items[1..$].filter!(i=> i<items[0] ).array),
        items[0..1],
        qs(items[1..$].filter!(i=> i >= items[0]).array)
      ).array;
	
    auto result = qs([10,9,5,4,8,7,-20]);
    assert( result.equal([-20,4,5,7,8,9,10]) );
    writeln("Result:", result );
}

First problem I found is "qs" must be splitted in 2 expressions: declaration and assignation, because declaration is not "effective" until all expression is compiled (compiler says qs doesn't exist when compiling the lambda body) .

* Is there any way to reduce the code(using lambdas) to one expression only?

  int[] delegate(int[]) qs = (int[] items) => items.length==0 .....

  Or better

  auto qs = (int[] items) => items.length==0 ?  ...

* Can the lambda be transformed to a template (using T instead "int") but avoiding function/return syntax?

This is an example using function

  template qs(T){
    T[] qs( T[] items ){
      return items.length==0
        ? items
        : chain(
          qs(items[1..$].filter!(i=> i<items[0] ).array),
          items[0..1],
          qs(items[1..$].filter!(i=> i >= items[0]).array)
        ).array;
    }
}


* I'm trying to use "ranges" to avoid the "array" conversion.  Do you figure out a way to express the lambda params and return as a Range to avoid converting to array?

On Friday, 27 November 2020 at 16:40:43 UTC, ddcovery wrote:
> ...
> * Can the lambda be transformed to a template (using T instead "int") but avoiding function/return syntax?
>
> This is an example using function
>
>   template qs(T){
>     T[] qs( T[] items ){
>       return items.length==0
>         ? items
>         : chain(
>           qs(items[1..$].filter!(i=> i<items[0] ).array),
>           items[0..1],
>           qs(items[1..$].filter!(i=> i >= items[0]).array)
>         ).array;
>     }
> }

I mean... transform this into a lambda expression:

    T[] qs(T)( T[] items ){
      return items.length==0
        ? items
        : chain(
          qs(items[1..$].filter!(i=> i<items[0] ).array),
          items[0..1],
          qs(items[1..$].filter!(i=> i >= items[0]).array)
        ).array;
    }

On 11/27/20 9:01 AM, ddcovery wrote:
> On Friday, 27 November 2020 at 16:40:43 UTC, ddcovery wrote:
>> ...
>> * Can the lambda be transformed to a template (using T instead "int") but avoiding function/return syntax?
>>
>> This is an example using function
>>
>>   template qs(T){
>>     T[] qs( T[] items ){
>>       return items.length==0
>>         ? items
>>         : chain(
>>           qs(items[1..$].filter!(i=> i<items[0] ).array),
>>           items[0..1],
>>           qs(items[1..$].filter!(i=> i >= items[0]).array)
>>         ).array;
>>     }
>> }
> 
> I mean... transform this into a lambda expression:
> 
>      T[] qs(T)( T[] items ){
>        return items.length==0
>          ? items
>          : chain(
>            qs(items[1..$].filter!(i=> i<items[0] ).array),
>            items[0..1],
>            qs(items[1..$].filter!(i=> i >= items[0]).array)
>          ).array;
>      }
> 

This has been done with the Y-combinator, where the lambda refers to itself as 'self':


https://github.com/gecko0307/atrium/blob/master/dlib/functional/combinators.d

There has been been other discussions on it on these forums.

Ali

On Friday, 27 November 2020 at 20:53:37 UTC, Ali Çehreli wrote:

> This has been done with the Y-combinator, where the lambda refers to itself as 'self':
>
>
> https://github.com/gecko0307/atrium/blob/master/dlib/functional/combinators.d
>
> There has been been other discussions on it on these forums.
>
> Ali

Thanks Ali.