无法在Lua中的多线程编程中注册信号处理程序

Question

我目前正在使用火炬来训练多线程深度学习网，我需要注册一个信号处理程序，这样当我按Docker Quickstart Terminal.app时我可以保存网络的权重。但问题是，我发现多线程程序只是忽略了信号，我完全不明白为什么......

见下面的代码

ctrl+c

和

-- threadTester.lua
local classic = require 'classic'
local threads = require 'threads'
threads.Threads.serialization('threads.sharedserialize')
local tds = require 'tds'
local threadTester = classic.class('threadTester')

function threadTester:_init(atomic)
    self.game = threads.Threads(1,
    function ()
        inner = atomic
    end)
    classic.strict(self)
end

function threadTester:play()
    self.game:addjob(function ()
        while true do
            if inner:get() < 0 then break end
        end
    end)
    -- do some stuff outside
end

return threadTester

在macOS 10.12，火炬7中，拨打local threads = require 'threads' threads.Threads.serialization('threads.sharedserialize') local tds = require 'tds' local signal = require 'posix.signal' local ctrlpool = threads.Threads(1, function () local tds = require 'tds' end) local atomic = tds.AtomicCounter() atomic:set(1) nThreads = 4 local ctrlPool = threads.Threads(1) ctrlPool:addjob(function () local signal = require 'posix.signal' signal.signal(signal.SIGINT, function(signum) print('\nSIGINT received') print('Ex(c)iting') atomic:set(-1) end) end) ctrlPool:synchronize() local gamePool = threads.Threads(nThreads, function () threadTester = require 'threadTester' player = threadTester(atomic) end) for i = 1, nThreads do gamePool:addjob(function () print(string.format("begin in thread %d", __threadid)) local status, err = xpcall(player.play, debug.traceback, player) if not status then print(string.format('%s', err)) os.exit(128) end end) end gamePool:synchronize() gamePool:terminate() ，然后按th test.lua，没有任何反应。 但如果我按以下方式更改ctrl+c

test.lua

一切都如预期的那样

local threads = require 'threads'
threads.Threads.serialization('threads.sharedserialize')
local tds = require 'tds'
local signal = require 'posix.signal'

-- local ctrlpool = threads.Threads(1, function ()
--  local tds = require 'tds'
-- end)

local atomic = tds.AtomicCounter()
atomic:set(1)
nThreads = 4
-- local ctrlPool = threads.Threads(1)
-- ctrlPool:addjob(function ()
    local signal = require 'posix.signal'
    signal.signal(signal.SIGINT, function(signum)
        print('\nSIGINT received')
        print('Ex(c)iting')
        atomic:set(-1)
    end)
-- end)

-- ctrlPool:synchronize()

local gamePool = threads.Threads(nThreads, function ()
    threadTester = require 'threadTester'
    player = threadTester(atomic)
end)

for i = 1, nThreads do
    gamePool:addjob(function ()
        print(string.format("begin in thread %d", __threadid))
        local status, err = xpcall(player.play, debug.traceback, player)
        if not status then
            print(string.format('%s', err))
            os.exit(128)
        end
    end)
end

gamePool:synchronize()
gamePool:terminate()

但如果我使用> th test.lua begin in thread 1 begin in thread 2 begin in thread 3 begin in thread 4 ^C SIGINT received Ex(c)iting

qlua

再次失败...... 我很困惑......我不知道为什么它在>qlua test.lua begin in thread 1 begin in thread 2 begin in thread 3 begin in thread 4 ^C^C^C^C^C^C^Z [3] + 15370 suspended qlua test.lua 中不起作用，我确实看到有人在ctrl池中注册信号处理程序......

如果我像这样更改qlua

threadTester.lua

它的功能更加奇怪（使用no-ctrl-pool local classic = require 'classic' local threads = require 'threads' threads.Threads.serialization('threads.sharedserialize') local tds = require 'tds' local threadTester = classic.class('threadTester') function threadTester:_init(atomic) self.game = threads.Threads(1, function () inner = atomic end) self.atomic = atomic classic.strict(self) end function threadTester:play() self.game:addjob(function () os.execute("sleep ".. 10) end) -- do some stuff outside while true do if self.atomic:get() < 0 then break end end end return threadTester ）

test.lua

似乎仅在th test.lua begin in thread 1 begin in thread 2 begin in thread 3 begin in thread 4 ^C^C^C^C^C SIGINT received Ex(c)iting 完成后才收到信号。