Spawn远程进程w / o通用文件系统 [英] Spawn remote process w/o common file system

问题描述

([e-mail])8> spawn([e-mail], tut, test, [hello, 5]).

我想在bar.del.com上生成一个没有文件系统访问foo的进程。 hyd.com（从我产生的过程），运行模块tut的子程序test。

I want to spawn a process on bar.del.com which has no file system access to foo.hyd.com (from where I am spawning the process), running subroutine "test" of module "tut".

有没有办法这样做，w / o提供[e-mail]与编译的tut模块文件？

Is there a way to do so, w/o providing the [e-mail] with the compiled "tut" module file?

推荐答案

您可以使用以下功能在远程节点加载模块，而不提供文件本身：

You can use the following function to load a module at remote node without providing the file itself:

load_module(Node, Module) ->
    {_Module, Bin, Filename} = code:get_object_code(Module), 
    rpc:call(Node, code, load_binary, [Module, Filename, Bin]).

如代码所示：load_binary / 3 文件名参数仅用于跟踪模块的路径，并且其指向的文件不被本地node_server使用。

As noted in code:load_binary/3 Filename argument is used only to track the path to module and the file it points to is not used by local node_server.

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