lua synchronous call C asynchronous function - c

Question

c call lua with the following code
if ( lua_pcallk(L, 0, LUA_MULTRET, 0, ctx, pcallk_continue) != 0 )
{}
the lua code is
local coroutine = coroutine
co = coroutine.create( function (...)
c.call(...)
-- [run here]
end )
c.call() is an asynchronous c function(it call lua_yieldk() at the end of the function, util the network recv the data and call lua_resume() )
return lua_yieldk(L, 0, 0, yield_continue);
it run to yield_continue function, but not run to the line of "[run here]", why?
the reference URL: http://www.lua.org/manual/5.2/manual.html#4.7

Answer 1

This is main idea with dispatcher.
Worker thread start async operation and switch to dispatcher thread.
Dispatcher thread polls async operations and when they done switch to
worker thread.
-- basic async operation queue
local ASYNC_RECV = {}
-- Put operation to queue and wait
local function recv()
local co = coroutine.running()
-- here we start async operation
ASYNC_RECV[co] = true
-- here we switch to dispatcher
return coroutine.yield()
end
coroutine.wrap(function()
for i = 1, 10 do print(recv()) end
end)()
-- main Loop
-- here we poll all async operation and
-- resume according coroutines
while true do
-- we have no async operation
if not next(ASYNC_RECV) then break end
for co in pairs(ASYNC_RECV) do
ASYNC_RECV[co] = nil -- mark operation is done
coroutine.resume(co, "some string") -- send result to coroutine
end
end
And here work example with libuv as dispatcher
local uv = require "lluv"
local function sleep(timeout)
local co = coroutine.running()
-- here we start async operation
uv.timer():start(timeout, function(timer)
timer:close() -- just close unused timer.
-- dispatcher(libuv) call this callback and
-- callback switch to coroutine
coroutine.resume(co)
end)
coroutine.yield()
end
coroutine.wrap(function()
for i = 1, 10 do
print("Tick")
sleep(1000)
end
end)()
uv.run()