I just replace this line

    nbCrefPrevious = Util.if_(List.isEmpty(indexList),1,List.last(indexList));

by this :   nbCrefPrevious = count4(indexList);
with
public function count4
  input  list<Integer> indexList;
  output Integer nbCrefPrevious;
algorithm
nbCrefPrevious := matchcontinue(indexList)
  case({}) then 1;
  case(_) then List.last(indexList);
end matchcontinue;
end count4;

But I does not run correctly ,  something happens in count3 method !!

Alain

Nowly, I don't consider <list<list<String> so that my problem is written like that :

But I have alaways a mistake for the line which computes the elemPosition1,


Regards

Alain

//-------------------------------------
public function countStrCref "Specific count about StrCref."
  input  list<String> inList;
  input  list<Integer> indexList;
  output list<Integer> outList;
algorithm
  outList := matchcontinue(inList,indexList)
    local
       String elem;
       list<String> tail;

       Integer nbCref,nbCrefPrevious,elemPosition1;
       list<Integer> nbCrefIndex;
       list<String> inList2;
    case({},_)
       then {};
    case(elem :: tail,_)
     equation
       nbCrefPrevious = getCountBefore(indexList);

       print("3\n");
       elemPosition1 = List.position(elem,tail);

       print("4\n");
       elemPosition1 = elemPosition1 + 1;

       //inList2  = Util.if_(elemPosition1>listLength(tail),{},List.lastN(tail,elemPosition1));

       inList2  = Util.if_(stringEqual(elem,List.last(tail)),{},List.lastN(tail,elemPosition1));

       print("4.1\n");
       nbCrefIndex = count3(inList2,elem,{});

       nbCref = count2(nbCrefPrevious,nbCrefIndex);
       print("5\n");
     then
        countStrCref(tail,listAppend(indexList,{nbCref}));
  end matchcontinue;
end countStrCref;
public function count2
  input Integer nbCrefPrevious;
  input list<Integer> nbCrefIndex;
  output Integer nbCref;
algorithm
   nbCref := matchcontinue(nbCrefPrevious,nbCrefIndex)
      local
         Integer n,nout;
    case(n,{})
      then 1;
    case(n,_)
      equation
       print("4.2\n");
       nout = Util.if_(n > 1,-n+1,
                 Util.if_(n < 0 and n != -1,n+1,
                      Util.if_(n == 1 or n == -1,List.reduce(nbCrefIndex,intAdd), 1)));
       print("4.3\n");
      then
         nout;
    end matchcontinue;
end count2;

public function count3
   input  list<String> inList;
   input  String strCrefToCompare;
   input  list<Integer> nbCrefLst;
   output list<Integer> nbCref;
algorithm
  nbCref := matchcontinue(inList,strCrefToCompare,nbCrefLst)
     local
        Integer nbFirstCref;
        list<String> strCrefIndxLst;
        String  strCrefIndx,elem;
        list<String> tail;
    case({},_,_)
        then {};
    case(elem :: tail,_,_)
     equation
       print("4.1.1\n");
       //strCrefIndx = listGet(elem,1);
       strCrefIndxLst = Util.stringSplitAtChar(elem,".");
       strCrefIndx = listGet(strCrefIndxLst,1);
       print("4.1.2\n");
       nbFirstCref = Util.if_(stringEqual(strCrefToCompare,strCrefIndx),1,0);
       print("4.1.3\n");
     then count3(tail,strCrefToCompare,listAppend(nbCrefLst,{nbFirstCref}));
end matchcontinue;
end count3;
/-------------------------------------
public function getCountBefore
  input  list<Integer> indexList;
  output Integer nbCrefPrevious;
algorithm
nbCrefPrevious := matchcontinue(indexList)
  case({}) then 1;
  case(_)  then List.last(indexList);
end matchcontinue;
end getCountBefore;