[Erl_question16] Why should I use MFA instead of fun ()-> end?

MFA: Module function arguments.

First, you need to know the differences between module: func (ARGs) and func (ARGs?

If you are interested in the details, you can learn about: http://www.cnblogs.com/zhongwencool/p/erlang_hot_code.html

In short: Erlang functions have the difference between local call and external call. Local call is called in the defined module and can be called directly: func (ARGs ); external call is an explicit call that uses module: func (ARGs) to call or import other modules.

When two versions of the same module are loaded, all local calls can work in the current version. However, external call can only be called to the latest version!

1. Fun scenarios:

1.1 first start one shell terminal named test1:

erl -name test1

1.2 test1 loading module mfa_test.erl

-module(mfa_fun).-export([cal/2,func/0]).cal(X,Y) ->     X+Y.func() ->       fun(X,Y) -> cal(X,Y) end.

1.3 input in shell:

>Func = mfa_fun:func().#Fun<mfa_fun.1.52492151>>Func(1,2).3

2. MFA scenarios:

Change the cal function in mfa_test.erl and re-load this module:

cal(X,Y) -> X*Y .

1.5 run again:

>Func(1,2).3

You can see the local version of the func running. If you change func to the following version:

func() ->       fun(X,Y) -> ?MODULE:cal(X,Y) end.

The result is the latest version of code that will be called forever. If you are interested, you can verify it by yourself. Why is the func variable unchanged, but the result has changed...

We can further consider:

We often use spawn (func) to create a new process. If we never upgrade, what will happen if it is a hot upgrade?

You cannot change the local call func, so it is best to use it normally:

Spawn (mod, funcname, argS) to create a process, which does not affect the upgrade.

FurtherAll local calls stored in the process cannot be upgraded. The best practice is to always use MFA!

 

Choice is more important than hard work. Be grateful in time and cherish life.