我不知道如何在模块中调用 local 函数，以便在代码更改后使用最新版本的代码。请参见以下示例：
要更改的函数是add/1。为了使用最新版本的函数，add/1的调用（第17行）应该是 * 完全限定函数调用 * {Pid, Z} -> Pid ! ?MODULE:add(Z)。当我尝试它时，得到如下结果：
1> c(test).
{ok,test}
2> test:start().
true
3> test:call(1).
2第22行变更为N + 2
4> c(test).
{ok,test}
5> test:call(1).
3第22行再次更改为N + 3
6> c(test).
{ok,test}
7> test:call(1).
** exception error: bad argument
in function test:call/1 (test.erl, line 10)为什么会出现此错误？
1条答案
按热度按时间dpiehjr41#
I believe that you need to eventually call the fully qualified version of the
loop/0function instead of theadd/1function in order to load and use the new module. The code loading mechanism is prepared to handle two running versions of a module at once, and your example withN+3is the third load of the module -- and the first versions is forcibly removed.Try instead this loop:
I've changed it to reload the newest version on next execution of the
loop/0.I believe more common is to use a
reloadmessage or similar that will directly call the main loop explicitly, to avoid the overhead of constantly reloading the module on every request.