Time consuming function
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- Subject: Time consuming function
- From: Eugen-Andrei Gavriloaie <shiretu@...>
- Date: 2008年3月31日 13:56:49 +0300
Hello,
Here is a little scenario that I want to use:
-- /tmp/test.lua --
function test()
print("Before")
a=cfunc(123)
print("After",a)
end
-- end tmp/test.lua --
cfunc is a function defined in C++ like this:
-- cut here --
int w_cfunc_call(lua_State *pLuaState)
{
//Get the parameters from pLuaState
.....
//I enqueue a request. I don't know when it will finish
//It will run in a separate thread. When it will be finished
//I will get signaled and I will do a lua_resume. Till then
//we must issue a lua_yield
enqueue_request(...);
lua_yield(pLuaState,1);
return 1;
}
....
lua_pushcfunction(pLuaState, w_cfunc_call);
lua_setglobal(pLuaState, "cfunc");
...
-- end cut --
Basically, when I do `a=cfunc(123)` it will take a looong time
(database access). I want to yield the current coroutine from C++
so the lua script writer won't feel anything. But When I run my app I
get '/tmp/test.lua: attempt to yield across metamethod/C-call boundary'
To further summarize the problem:
1. Have a c++ working thread that gets it's work chunk from a queue
2. Wrap all the internals in a c++ function callable from Lua
a. enqueue the work chunk
b. yield Lua
c. wait for the results from thread (we will get signaled)
d. resume Lua by invoking lua_resume
3. The writer of the Lua script must not know anything about the
internals of the call (we must not force him to make a yield/resume
from Lua)
Is this possible?