Finally I took Andrea's solution "check_elem (x:xs) = if x == e then (l2 ++ xs) else [x] ++ check_elem xs" I think it's easy to understand for me ( in my noob level), than the recursive one. I'm testing it and it's working really well. The other solutions are a little complicated for me, but I'm still trying to undestand them. Thanks!
Andrea Rossato wrote: > > On Mon, Sep 18, 2006 at 12:25:21PM +0100, Neil Mitchell wrote: >> Why not: >> > check_elem (x:xs) = if x == e then (l2 ++ xs) else x : check_elem xs >> >> Thanks > > Thank you! > Lists are my personal nightmare...;-) > > Andrea > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > > -- View this message in context: http://www.nabble.com/Problems-interpreting-tf2290155.html#a6362822 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
