What's the difference in functionality between these two blocks of code?

Question

First:

-module(some_mod).
-compile(export_all).

some_fun() ->
    fun f/0.

f() ->
    ok.

Second:

-module(some_mod).
-compile(export_all).

some_fun() ->
    fun ?MODULE:f/0.

f() ->
    ok.

I encountered this change during a hot code upgrade. What is the the difference between fun ?MODULE:f/0 and fun f/0?

Answer 1

From Erlang documentation:

• A fun created by fun M:F/A is called an external fun. Calling it will always call the function F with arity A in the latest code for module M. Notice that module M does not even need to be loaded when the fun fun M:F/A is created.

• All other funs are called local fun. When a local fun is called, the same version of the code that created the fun is called (even if a newer version of the module has been loaded).

They have different behaviours in code upgrading as the documentation says. Your first module uses a local function (fun f/0) and the second one uses an external function (fun ?MODULE:f/0 which in preprocessing replaced with fun some_mod:f/0).

So if you upgrade your first module (which uses local function), the processes that are using some_fun function, don't use the newer version. But if you upgrade the second module (which uses external function), the latest version of code will be called whenever some_fun is called from inside processes which were spawned even before the loading of new version.

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Notice: There can be just two versions of a module, old and new. If a third version of the module is loaded, the code server removes (purges) the old code and any processes lingering in it is terminated.