Not a good solution, it just substitutes the first occurrence of the item in the list. I'll try the others
Carajillu wrote: > > 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 >> [email protected] >> http://www.haskell.org/mailman/listinfo/haskell-cafe >> >> > > -- View this message in context: http://www.nabble.com/Problems-interpreting-tf2290155.html#a6362912 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
