Getting to know who peer reviews a Sci-Writing assignment

peerReview.erl

-module(peer).
-export ([draw/0]).

draw()->
    R = random:uniform(100),
    io:format ("~w",[r]),
    if  
        R < 33 ->
            io:format ("Vasia peer reviews Erisa's S. Writing");
        true ->
            if  
            R < 66 ->
                io:format ("Lalith peer reviews Erisa's S. Writing");
            true ->
                io:format ("Mariano peer reviews Erisa's S. Writing")
            end 
    end.

Por ¿ qué

Para recordar un poco Erlang !

…

On Mon, Oct 17, 2011 at 1:10 AM, Nicholas Rutherford ***@***.*** wrote: Por ¿ qué ## Reply to this email directly or view it on GitHub: https://gist.github.com/1291190

Hm... Usted está loco! As an improvement though, may I suggest you remove the if-clauses and use pattern matching instead ;)

how could you pattern match with conditions? wow, I have really lost my Erlang.
Let's say now there are only two possibilities to make it simple , I could do:
draw() ->
R = random:uniform(100),
if
R < 50 ->
test ! {smallerThan33, R} ,
true ->
test ! {biggerThan33, R}
end
And do the pattern matching and printouts in the test function.
What other way do you propose?

I'd suggest case/switch, since that allows simple predicate tests which you can't put in function definitions, but it'll look quite similar.

One way could be:

-module(peer).
-export ([draw/0]).

draw()->
    R = random:uniform(100),
    io:format ("~p peer reviews Erisa's S. Writing~n", [draw(R)]).

draw(R) when R < 33 ->
    'Vasia'; 
draw(R) when R < 66 ->
    'Lalith';
draw(_R) ->
    'Mariano'.

What about using eval:

-module(peer).
-export ([draw/0]).

draw() -> os:cmd("ruby -e 'print %w(Vasia Lalith Mariano).choice'").

Hahah! That's avoiding the problem.