问题描述
我正在尝试构建一个非常简单的屏障同步服务器，该服务器最初会被喂以许多与之通信的进程.进程完成后，它将收到带有该进程的Pid的消息，并保留要执行此操作的每个进程的列表.当屏障达到零(所有进程都已发送消息)时，我的服务器需要向其中每个发送一条消息(我使用 [Pid | ProcList] 作为我的pids列表).
I am trying to build a very simple barrier-synchronization server, where the server is initially fed a number of processes that will be communicating with it. When a process is done, it receives a message with that process' Pid, and it keeps a list of every process to do so. When the barrier reaches zero (all processes have sent messages), my server needs to send a message to each of these (I am using [Pid | ProcList] as my list of pids).
我尝试使用辅助函数无济于事，列表理解使我陷入无限循环，因此我正在研究如何使用list:foreach来解决这一问题.我对函数式编程相当陌生，但是据我了解，此foreach需要接收列表以及lambda微积分函数，以将消息发送到列表中的每个节点.由于！"的中缀性质，我尚未找到一种在不引起语法错误的情况下进行此操作的方法.
I have tried using a helper function to no avail, list comprehensions keep me in an infinite loop, and as such I am looking into how to use lists:foreach to take care of this.I am fairly new to functional programming, but from what I understand, this foreach needs to take in the list as well as a lambda-calculus function to send a message to each node in the list. Due to the infix nature of "!", I have yet to find a way to do this without causing syntax errors.
推荐答案
如何在列表理解中实现无限循环?我必须说，这非常具有挑战性.试试这个:
How you've made infinite loop in list comprehension? I must say, that's quite challenging. Try this:
Message = % broadcast message goes here
ListOfPids = % list of recipients
[Pid ! Message || Pid <- ListOfPids].
如果要使用 foreach ，则它将一个参数函数作为第一个参数，因此需要先包装 send ，因为它是两个参数函数.
If you want to use foreach, than it takes one argument function as first argument, so need to wrap send first, as it is two argument function.
Message = % broadcast message goes here
ListOfPids = % list of recipients
Fun = fun (Pid) -> Pid ! Message end,
lists:foreach(Fun, ListOfPids).
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