是否可以存储像
这样的模板化类template <typename rtn, typename arg>
class BufferAccessor {
public:
int ThreadID;
virtual rtn do_work(arg) = 0;
};
BufferAccessor<void,int> access1;
BufferAccessor<int,void> access2;
像 vector 或列表一样在同一个容器中
编辑: 这样做的目的是我试图制作一个循环缓冲区,其中想要使用缓冲区的对象需要向缓冲区注册。缓冲区将存储一个 boost::shared_ptr 到访问器对象,并生成一个回调到那里的函数,这些函数将数据推送到缓冲区或从缓冲区提取数据。回调将用于我创建的类似于线程池的通用线程工作函数,事实上它们需要访问共享内存对象。下面是我输入的一些代码,可能有助于说明我正在尝试做的事情,但它还没有被编译,这也是我第一次使用绑定(bind)、函数、多线程
typedef boost::function<BUF_QObj (void)> CallbackT_pro;
typedef boost::function<void (BUF_QObj)> CallbackT_con;
typedef boost::shared_ptr<BufferAccessor> buf_ptr;
// Register the worker object
int register_consumer(BufferAccesser &accessor) {
mRegCons[mNumConsumers] = buf_ptr(accessor);
return ++mNumConsumers;
}
int register_producer(BufferAccesser &accessor) {
mRegPros[mNumProducers] = buf_ptr(accessor);
return ++mNumProducers;
}
// Dispatch consumer threads
for(;x<mNumConsumers; ++x) {
CallBack_Tcon callback_con = boost::bind(&BufferAccessor::do_work, mRegCons[x]);
tw = new boost:thread(boost::bind(&RT_ActiveCircularBuffer::consumerWorker, this, callback_con));
consumers.add(tw);
}
// Dispatch producer threads
for(x=0;x<mNumProducers; ++x) {
CallBack_Tpro callback_pro = boost::bind(&BufferAccessor::do_work, mRegPros[x], _1);
tw = new boost:thread(boost::bind(&RT_ActiveCircularBuffer::producerWorker, this, callback_pro));
producers.add(tw);
}
// Thread Template Workers - Consumer
void consumerWorker(CallbackT_con worker) {
struct BUF_QObj *qData;
while(!mRun)
cond.wait(mLock);
while(!mTerminate) {
// Set interruption point so that thread can be interrupted
boost::thread::interruption_point();
{ // Code Block
boost::mutex::scoped_lock lock(mLock);
if(buf.empty()) {
cond.wait(mLock)
qData = mBuf.front();
mBuf.pop_front(); // remove the front element
} // End Code Block
worker(qData); // Process data
// Sleep that thread for 1 uSec
boost::thread::sleep(boost::posix_time::nanoseconds(1000));
} // End of while loop
}
// Thread Template Workers - Producer
void producerWorker(CallbackT_pro worker) {
struct BUF_QObj *qData;
boost::thread::sleep(boost::posix_time::nanoseconds(1000));
while(!mRun)
cond.wait(mLock);
while(!mTerminate) {
// Set interruption point so that thread can be interrupted
boost::thread::interruption_point();
qData = worker(); // get data to be processed
{ // Code Block
boost::mutex::scoped_lock lock(mLock);
buf.push_back(qData);
cond.notify_one(mLock);
} // End Code Block
// Sleep that thread for 1 uSec
boost::thread::sleep(boost::posix_time::nanoseconds(1000));
} // End of while loop
}
最佳答案
不,不是,因为 STL 容器是同质的,access1 和 access2 具有完全不同的不相关类型。但是您可以将类 BufferAccessor 设为非模板,而将 do-work 成员设为模板,如下所示:
class BufferAccessor
{
template<class R, class A>
R doWork(A arg) {...}
};
在这种情况下,您可以将 BufferAccessors 存储在容器中,但不能将成员模板函数设为虚拟。
关于c++ - 在容器中存储模板化对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3962615/