[erlang-questions] Erlang again!

Jilani Khaldi <>
Fri Jul 23 13:15:10 CEST 2010


> This says
>      for each X1 in L1 {
>          for each X2 in L2 {
>              include X1 - X2 in the result
>          }
>      }
>
>   ** sub(Xs, Ys) ->  [X-Y || X<- Xs&&  Y<- Ys]
I find this last one (not yet implemented in Erlang) the more natural 
and expressive solution to the question.

-- 
Jilani KHALDI
---------------------
http://www.iiiaugusta.com



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