[erlang-questions] passing a function to lists:map

[Raimo Niskanen]

On Mon, Mar 22, 2010 at 01:01:04PM +0100, Jeroen Koops wrote:
> Really now? That goes to show, teaching =:= learning !
> 
> On Mon, Mar 22, 2010 at 12:57 PM, Bengt Kleberg
> <bengt.kleberg@REDACTED>wrote:
> 
> > Greetings,
> >
> > You do not need to export the function to be able to use it in a
> > fun func/arity
> > expression.

The export is not needed since the "fun Fun/1"
construct is essentially syntactical sugar for
"fun (Arg) -> Fun(Arg) end" like this:

    atomize(List) -> lists:map(fun(Arg) -> ta(Arg) end, List).

So you could do a macro approach to replace "fun Fun/1":

    -define(FUN1(Fun), fun (Arg) -> (Fun)(Arg) end).

    atomize(List) -> lists:map(?FUN1(ta), List).

> >
> >
> > ebngt
> >
> > On Mon, 2010-03-22 at 12:53 +0100, Jeroen Koops wrote:
> > > Hi Martin,
> > >
> > > use:
> > >
> > > -export([ ta/1 ]).
> > >
> > > and then:
> > >
> > > atomize(List) -> lists:map(fun ta/1, List).
> > >
> > > Regards,
> > >
> > >
> > >
> > > On Mon, Mar 22, 2010 at 12:49 PM, Martin DeMello <
> > martindemello@REDACTED>wrote:
> > >
> > > > ta(X) when is_list(X)    -> list_to_atom(X);
> > > > ta(X) when is_tuple(X)   -> tuple_to_list(X);
> > > > ta(X) when is_integer(X) -> X.
> > > >
> > > > % this doesn't work
> > > > atomize(List) -> lists:map(ta, List).
> > > >
> > > > % this does
> > > > atomize(List) -> lists:map(fun(X) -> ta(X) end, List).
> > > >
> > > > What's the difference? Have I just got the syntax wrong, or is there
> > > > some fundamental reason I need the second version?
> > > >
> > > > martin
> > > >
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> >
> >
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-- 

/ Raimo Niskanen, Erlang/OTP, Ericsson AB