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<  Erlang  ~  foldl function call itself
mccmike
Posted: Fri Mar 05, 2010 12:32 am Reply with quote
Joined: 16 Feb 2010 Posts: 3
Stepping thru a large deep list

Within this list are tuples that contain lists.

Is it possible to call the foldl fun from within the foldl fun when I encounter one of these tuples?

Trying to do this

Walk = fun(Rec, Acc) ->
io:format("~nWalking ~p",[Rec])

case is_tuple(Rec) of
true -> lists:foldl(Walk, [], (Rec tuple list)
false -> ok
end
end,

lists:foldl(Walk, [], Doc).
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bpuzon
Posted: Fri Mar 05, 2010 12:55 pm Reply with quote
User Joined: 05 Aug 2009 Posts: 23 Location: Cracow, Poland
You have to pass the function as the paramenter - like:

Code:

   Walk = fun(Rec, {Fun, Acc}) ->
         io:format("~nWalking ~p~n",[Rec]),
         case is_tuple(Rec) of
             true ->  {Fun, lists:foldl(Fun, {Fun, Acc}, (Rec tuple list))};
             false -> {Fun, ok}
         end
      end,
    lists:foldl(Walk, {Walk, []}, Doc)

_________________
Saludos,
Bartłomiej Puzoń
Erlang Solutions
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mccmike
Posted: Fri Mar 05, 2010 4:02 pm Reply with quote
Joined: 16 Feb 2010 Posts: 3
Thanks
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