我是Erlang的新手,正在尝试编写有界缓冲问题程序。它几乎可以正常工作,除了确保生产者不会过于遥远并覆盖未使用的数据。为了处理这个问题,我决定尝试在我的buffer()函数上设置保护程序,这样我就可以在缓冲区已满时使用一个没有接收的版本,在缓冲区为空时使用一个没有发送的版本,并且正常其余时间的版本。
我的问题是,无接收器版本的保护要求我知道代表缓冲区的数组的大小,这需要调用array:size/1
。显然,Erlang不允许在警卫中进行函数调用,这会阻止它工作。有没有办法在不更改缓冲区actor的函数声明的情况下解决这个问题?
%% buffer: array num num
%% A process that holds the shared buffer for the producers and consumers
buffer(Buf, NextWrite, NextRead) when NextWrite == NextRead ->
io:format(" * ~w, ~w, ~w~n", [array:to_list(Buf), NextRead, NextWrite]),
receive
{enqueue, Reply_Pid, Num} ->
io:format("~w: > ~w~n", [Reply_Pid, Num]),
buffer(array:set(NextWrite rem array:size(Buf), Num, Buf), NextWrite + 1, NextRead);
finish ->
io:format("finished printing~n")
end;
buffer(Buf, NextWrite, NextRead) when (NextWrite - NextRead) == array:size(Buf) ->
io:format(" * ~w, ~w, ~w~n", [array:to_list(Buf), NextRead, NextWrite]),
receive
{dequeue, Reply_Pid} ->
io:format("~w: < ~w~n", [Reply_Pid, array:get(NextRead rem array:size(Buf), Buf)]),
Reply_Pid ! {reply, array:get(NextRead rem array:size(Buf), Buf)},
buffer(Buf, NextWrite, NextRead + 1);
finish ->
io:format("finished printing~n")
end;
buffer(Buf, NextWrite, NextRead) ->
io:format(" * ~w, ~w, ~w~n", [array:to_list(Buf), NextRead, NextWrite]),
receive
{dequeue, Reply_Pid} ->
io:format("~w: < ~w~n", [Reply_Pid, array:get(NextRead rem array:size(Buf), Buf)]),
Reply_Pid ! {reply, array:get(NextRead rem array:size(Buf), Buf)},
buffer(Buf, NextWrite, NextRead + 1);
{enqueue, Reply_Pid, Num} ->
io:format("~w: > ~w~n", [Reply_Pid, Num]),
buffer(array:set(NextWrite rem array:size(Buf), Num, Buf), NextWrite + 1, NextRead);
finish ->
io:format("finished printing~n")
end.
答案 0 :(得分:13)
只有某些功能可用于后卫,请参阅Guard Sequences in the Erlang manual。您可以按照以下方式轻松完成所需操作:
buffer(Buf, NextWrite, NextRead) -> buffer(Buf, NextWrite, NextRead, array:size(Buf)).
buffer(Buf, NextWrite, NextRead, _) when NextWrite == NextRead ->
;
buffer(Buf, NextWrite, NextRead, BufSize) when (NextWrite - NextRead) == BufSize ->
;
buffer(Buf, NextWrite, NextRead, _) ->
.
答案 1 :(得分:1)
正如Geoff Reedy所提到的那样,警卫只允许使用少量BIFS。
但guardian解析变换库可用于调用守卫中的任何函数。