if ... then ... else ... does not exist in RML MetaModelica. You can use Util.if_(cond, if_exp, else_exp), but it always evaluates both expressions.

ha, it works like a charm! Thanks.  I am pretty new to MetaModelica. Thanks a lot for helping me out on this. 

Is there any documentation about the grammar of RML? I checked the website but I dont know which one has a list of all statement available to use.

Thanks !

OpenModelicaMetaProgramming.pdf is the only document I know of. It covers the language, but the RML implementation is sometimes a bit different. For example the document states that listGet() is indexed from 1, but in RML MetaModelica we instead have listNth(), which is indexed from 0.

The appendix has both grammar and the builtin functions.

There is an OpenModelica development course here:
http://www.ida.liu.se/~pelab/modelica/O … ersCourse/
You can look at the slides and do the exercises to get more into it.

Cheers,
Adrian Pop/

And if you have trouble getting the exercises running, updated ones exist in subversion (trunk/testsuite/meta/MetaModelicaDev/).

Thank you all so much for the quick reply !  I have one more question regarding value comparison.  Thanks in advance.

Here is a function I wrote to traverse the incidence matrix and find the list of variables that a specific equation depends on.  inInteger is the equation number, and lst is the incidence matrix. counter is for checking whether this is the dependence we are interested.  The output should be a list of integer(variable number).
protected function findDep
  input Integer inInteger;
  input list<list<Integer>> lst;
  input Integer counter;
  output list<Integer> dep_list;
algorithm
  dep_list :=
  matchcontinue(inInteger, lst, counter)
    local
      Integer eqn;
      list<list<Integer>> lst_1;
      list<Integer> dep;
    case (_, {}, _) then {};
    case (_, (dep::lst_1), _)
      equation
           print(" in = ");
    print(intString(inInteger));
    print("     counter = ");
    print(intString(counter));
    print("\t");
        dep_list = Util.if_((inInteger ==counter), dep, findDep(inInteger, lst_1, counter+1));
      then dep_list;
  end matchcontinue;
end findDep;

The problem is, I dont know if Util.if_ is working as I expected, because the output is :
in = 10  counter = 1
in = 10  counter = 2
in = 10  counter = 3
in = 10  counter = 4
in = 10  counter = 5
in = 10  counter = 6
in = 10  counter = 7
in = 10  counter = 8
in = 10  counter = 9
in = 10  counter = 10
in = 10  counter = 11
in = 10  counter = 12
....

If the conditional branch really works, shouldn't it stop recursion when counter=10? I may have made some stupid mistake here since I am totally new to this.  Please please let me know if I am doing anything wrong.

Thanks for your time !

Yun

No. That's what I was referring to earlier. It will always evaluate both branches of the if-expression, which means you can't use it to break loops (it will try the other branch anyway).

Instead, start two cases like this:

case (_, (dep::lst_1), _) 

      equation

         true = inInteger == counter;

case (_, (dep::lst_1), _) 

      equation

         false = inInteger == counter;

I see. That explains a lot. Sorry I didn't understand what you meant by "evaluating both expressions".

In this case, if I want to break loops, shall I write program like:
case (_, (dep::lst_1), _)
      equation
         true = inInteger == counter;
        dep_list = Util.if_(true, dep, {});
      then dep_list;
case (_, (dep::lst_1), _)
      equation
         false = inInteger == counter;
         dep_list = Util.if_(false, {}, findDep(inInteger, lst_1, counter+1));
      then dep_list;

Yes. Since you cannot match with guards in MetaModelica, you need to break the loop like that.

Thanks, I dont know if I am doing anything else wrong: after I applied the  code above, the program always goes to the "false"branch in the first case.  The reason is that the fist case is taken, then the If in the fist case is evaluated. Given that the condition is false, the function always returns {}.....

In another word, instead of breaking the loop , the code above breaks the loop in its first iteration always because the first If is always evaluated before the second.

I am bit confused at this point...... How do I break the loop when inIntger==counter?

Here is the output  I got:
{10}
in = 10  counter = 1
finished
{8}
in = 8  counter = 1
finished

Gee,  it works ! Thank you very much !  Now I can write similar codes myself. Thanks!

Hi

There is actually a document for RML
look under
http://www.ida.liu.se/~pelab/rml/rml-ft … 060314.pdf
or
http://www.ida.liu.se/~pelab/rml/

Best Regards
Mohsen