[erlang-questions] Why it is a illegal guard expression?

Michael Richter ttmrichter@REDACTED
Wed Feb 24 04:48:41 CET 2010

There is a very limited list of which functions can be used in guard
expressions.  From
http://www.erlang.org/doc/reference_manual/expressions.html#id2273702 we

The set of valid guard expressions (sometimes called guard tests) is a
> subset of the set of valid Erlang expressions. The reason for restricting
> the set of valid expressions is that evaluation of a guard expression must
> be guaranteed to be free of side effects.

We also find a list of functions which can be used in guard expressions

On 24 February 2010 11:42, 钱晓明 <kyleqian@REDACTED> wrote:

> Hi, I am a newer of erlang. I wrote a module to find all files(no
> directory)
> under a directory, including its subdirectory:
> -module(lib_test).
> -export([list_files/1]).
> list_files(Path)->
>    {ok, FileList} = file:list_dir(Path),
>    process_filelist(FileList, []).
> process_filelist([File|FileList], Files)->
>    if
>       filelib:is_dir(File) ->
>        {ok, NewFileList} = file:list_dir(File),
>        process_filelist(NewFileList, Files);
>       true ->
>        process_filelist(FileList, [File|Files])
>    end;
> process_filelist([], Files) ->
>    Files.
> But I fail to compile this file, because " filelib:is_dir(File) ->" is
> illegal guard expression. So Why it is illegal? What should I write?
> Thanks for your help!

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