[erlang-questions] What is the fastest way to check if a function is defined in a module?

[Tristan Sloughter]

Ooh, that looks similar, but better, to what we did in Erleans:

    erlang:function_exported(Module, module_info, 0) orelse
    code:ensure_loaded(Module),    case erlang:function_exported(Module, FunctionName, Arity) of
        true ->
            erlang:apply(Module, FunctionName, Args);
        false ->
            Default
    end.

I guess we should likely switch to `erlang:module_loaded(M)` instead of
checking if `module_info` is exported. Thanks!
Would be nice to have it patched though :)

--
  Tristan Sloughter
  "I am not a crackpot" - Abe Simpson
  t@REDACTED


On Tue, Mar 6, 2018, at 12:23 PM, Richard Carlsson wrote:
> I was actually considering submitting such a patch a while back, but
> one argument against making that short-cut could be that operations
> against the code server should be serialized. On the other hand I
> haven't yet come up with an example where the short-cut would make a
> difference that couldn't already happen due to processor scheduling.> 
> 
>         /Richard
> 
> 2018-03-06 16:40 GMT+01:00 Michał Muskała <michal@REDACTED>:
>> 
>> On 6 Mar 2018, 16:36 +0100, Richard Carlsson
>> <carlsson.richard@REDACTED>, wrote:>> 
>> 
>>> 
>>> is_exported(M, F, A) ->
>>>   case erlang:module_loaded(M) of
>>>     false -> code:ensure_loaded(M);
>>>     true -> ok
>>>   end,
>>>   erlang:function_exported(M, F, A).
>>> 
>>> (code:ensure_loaded() is slow compared to the fast call to
>>> erlang:module_loaded(), even if the module is already in memory).>>>  
>> 
>> I was recently wondering about this. Is there a reason
>> code:ensure_loaded() does not short circuit using
>> erlang:module_loaded()? It could probably even skip the request to
>> the code server entirely in case the module is already loaded.>> 
>> 
>> Michał.
>> 
>> 
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