我的Lua脚本中有一种回调函数，我想从C ++端的不同线程调用（每秒0-100次）。到目前为止，它基本上可以工作，但是一旦我在很短的时间内多次调用它就会导致程序崩溃，从而导致错误，例如：
-As????ion failed: 0, file ...LuaFunction.h, line 146或this one（完全随机）
我认为这会发生，当它在完成另一个任务之前从C ++端调用它。对我来说最明显的事情是（在lua函数调用期间互斥锁定所有线程）根本没有帮助。 ：/
如果我每2秒只调用一次Lua函数，那么我根本就没有任何错误（好吧，直到清理部分，如果达到这一点，它将崩溃而没有特定的错误）。
这是我的代码（我试图尽可能地裁剪和简化我的代码，并添加了很多评论）：
#include "stdafx.hpp"
#include <pthread.h> //for multithreading
#include <windows.h>
#include <iostream>
#include <map>
using namespace std;
unsigned int maxThreads = 100;
map<unsigned int, pthread_t> threads;
map<unsigned int, bool> threadsState;
pthread_mutex_t mutex; //to lock the pthreads (to keep printing from overlapping etc)
LuaPlus::LuaState* pState = LuaPlus::LuaState::Create( true ); //initialize LuaPlus
LuaPlus::LuaObject globals = pState->GetGlobals();
struct argumentStruct { //to pass multiple arguments to the function called when starting a pthread
unsigned int threadId;
int a;
int b;
};
map<unsigned int, struct argumentStruct> argumentMap; //we store the arguments of active threads in here
void *ThreadFunction(void *arguments) { //will be called for every pthread we're going to create
struct argumentStruct*args = (struct argumentStruct*)arguments; //get the arrgument struct
int threadId = args->threadId; //get variables for each struct field
int a = args->a;
int b = args->b;
Sleep(3000); //since this is a very simplified version of my actual project
int c = a+b;
pthread_mutex_lock(&mutex); //lock pthreads for the next lines
LuaPlus::LuaFunction<int> CPP_OnMyEvent = pState->GetGlobal("LUA_OnMyEvent"); //get the Lua callback function to call on the C++ side
CPP_OnMyEvent(a,b,c); //call to our lua-callback function
pthread_mutex_unlock(&mutex); //unlock pthreads
threadsState[threadId] = false; //mark the thread as finished/ready to get overwritten by a new one
return NULL;
}
bool AddThread(int a, int b) {
for (;;) {
if (threads.size() < maxThreads) { //if our array of threads isn't full yet, create a new thread
int id = threads.size();
argumentMap[id].threadId = threads.size();
argumentMap[id].a = a;
argumentMap[id].b = b;
threadsState[id] = true; //mark the thread as existing/running
pthread_create(&threads[id], NULL, &ThreadFunction, (void *)&argumentMap[id]);
return true;
} else {
unsigned int id;
for (auto thread=threads.begin(); thread!=threads.end(); ++thread) {
id = thread->first;
if(!threadsState[id]) { //if thread with id "id" has finished, create a new thread on it's pthread_t
argumentMap[id].threadId = id;
argumentMap[id].a = a;
argumentMap[id].b = b;
threadsState[id] = true; //mark the thread as existing/running
pthread_join(threads[id], NULL);
pthread_create(&threads[id], NULL, &ThreadFunction, (void *)&argumentMap[id]);
return true;
}
}
}
}
return false;
}
int main() {
pthread_mutex_init(&mutex, NULL); //initialize the mutex
//LuaPlus::LuaState* pState = LuaPlus::LuaState::Create( true ); //we already initialized this globally
//LuaPlus::LuaObject globals = pState->GetGlobals();
//pState->DoString("function LUA_OnMyEvent(arg1,arg2) print(arg1..arg2) end"); //it's already in main.lua
globals.RegisterDirect("AddThread", AddThread);
char pPath[ MAX_PATH ];
GetCurrentDirectory(MAX_PATH,pPath);
strcat_s(pPath,MAX_PATH,"\\main.lua");
if( pState->DoFile(pPath) ) { //run our main.lua script which contains the callback function that will run a print
if( pState->GetTop() == 1 )
std::cout << "An error occured: " << pState->CheckString(1) << std::endl;
}
for (auto thread=threads.begin(); thread!=threads.end(); ++thread) { //wait for threads to finish
unsigned int id = thread->first;
if(threadsState[id])
pthread_join(threads[id], NULL);
}
//clean up
LuaPlus::LuaState::Destroy( pState );
pState = nullptr;
pthread_mutex_destroy(&mutex);
getchar(); //keep console from closing
return 0;
}
main.lua
function LUA_OnMyEvent(a,b,c)
print(a.."+"..b.."="..c)
end
for i=1, 999, 1 do
AddThread(i,i*2)
end
答案 0 :(得分：1)
我不知道Lua足以在Lua方面为您提供解决方案，但是这个问题的观点可能会帮助您实现这一目标。
当您从Lua调用AddThread（）时，会发生类似这样的事情：
1. LuaState allocations
2. AddThread() execution
3. LuaState unwinding
在ThreadFunction（）...
上A. Mutex lock
B. LuaState allocations
C. LUA_OnMyEvent() execution
D. LuaState unwinding
E. Mutex Unlock
AddThread没有互斥锁控制，因此竞争条件可能发生在1/3和B / D之间。 但是，将互斥锁添加到AddThread并不能解决问题，因为它仍然会在1到3之间运行。
如果仅在程序初始化时调用AddThread（），则可以阻止所有线程直到初始化完成。如果在程序执行期间经常调用它，那么我将从单独的LuaState进行调用。
[编辑]第二个想法：使用生产者/消费者方法。然后C ++线程不再需要运行Lua代码。
C ++建议：
//-- start Task.h --
struct Task{
static list<Task*> runningTasks;
static list<Task*> doneTasks;
static pthread_mutex_t mutex;
list<Task*>::iterator iterator;
virtual ~Task(){}
bool start(){
pthread_mutex_lock(&mutex);
bool hasSpace = runningTasks.size() < 100;
if(hasSpace){
runningTasks.push_front(this);
iterator = runningTasks.begin();
pthread_t unusedID;
pthread_create(&unusedID, NULL, Task::threadBody, this);
}
pthread_mutex_unlock(&mutex);
return hasSpace;
}
virtual void run() = 0;
virtual void processResults() = 0;
protected:
void finish(){
pthread_mutex_lock(&mutex);
runningTasks.erase(iterator);
doneTasks.push_front(this);
pthread_mutex_unlock(&mutex);
}
static void* threadBody(void* instance){
Task* task = static_cast<Task*>(instance);
task->run();
task->finish();
return NULL;
}
};
//-- end Task.h --
//-- start Task.cpp --
//Instantiate Task's static attributes
pthread_mutex_t Task::mutex;
list<Task*> Task::runningTasks;
list<Task*> Task::doneTasks;
//-- end Task.cpp --
struct SumTask: public Task{
int a, b, c;
void run(){
Sleep(3000);
c = a+b;
}
void processResults(){
LuaPlus::LuaFunction<int> CPP_OnMyEvent = pState->GetGlobal("LUA_OnMyEvent");
CPP_OnMyEvent(a,b,c);
}
}
//functions called by Lua
bool runSumTask(int a, int b){
SumTask task* = new SumTask();
task->a = a; task->b = b;
bool ok = task->start();
if(!ok)
delete task;
return ok;
}
int gatherResults(){
pthread_mutex_lock(&Task::mutex);
int totalResults = Task::doneTasks.size();
while(Task::doneTasks.size() > 0){
Task* t = Task::doneTasks.front();
Task::doneTasks.pop_front();
t->processResults();
delete t;
}
pthread_mutex_unlock(&Task::mutex);
return totalResults;
}
int main() {
//Must initialize/destroy Task::mutex
pthread_mutex_init(&Task::mutex, NULL);
//...
pthread_mutex_destroy(&Task::mutex);
}
Lua代码：
function LUA_OnMyEvent(a,b,c)
print(a.."+"..b.."="..c)
end
local totalRunning = 0;
for i=1, 999, 1 do
if (runSumTask(i,i*2))
totalRunning = totalRunning + 1;
totalRunning -= gatherResults();
end
while(totalRunning > 0) do
totalRunning -= gatherResults();
mySleepFunction(...);
end