boost.thread死锁和自删除

boost::thread_group 创建(使用) thread_group::create_thread() )并分派线程。为了限制 最大线程数,在每个线程结束时,我移除线程 从 thread_group 然后删除线程本身(这样我就可以 决定是否需要创建新线程)。不管它挂在哪里 在创建和删除最后一个线程之间(比如

我的问题是:

• 这
• 为什么我的代码挂起?

以下是相关代码:

//1-创建和分派线程的代码

 { 
        //mutex for map<thread_id, thread*> operations 
        boost::mutex::scoped_lock lk(m_mutex_for_ptr); 

        // create a thread for this->f(duplicate_hashes) 
        boost::thread* p = m_thread_group.create_thread(boost::bind( 
            &detectiveT<equal_predicate>::f, 
            this, 
            duplicate_hashes 
            )); 

        // save the <thread_id,thread pointer> map for later lookup & deletion 
        m_thread_ptrs.insert(make_pair(p->get_id(), p)); 

        // log to console for debug 
        cout << "thread created: " 
            << p->get_id() << ", " 
            << m_thread_group.size() << ", " m_thread_ptrs.size() << 
"\n";     
    }   

//2-线程执行的代码

void f(list<map_iterator_type>& l) 
{ 
    Do_something(l);    
    boost::this_thread::at_thread_exit(boost::bind( 
        &detectiveT<equal_predicate>::remove_this_thread, 
        this 
        ));                     
} 

//3-删除线程本身的代码

void remove_this_thread() 
{ 

    { 
        //mutex for map<thread_id, thread*> operations 
        boost::mutex::scoped_lock lk(m_mutex_for_ptr);                   
        boost::thread::id this_id(boost::this_thread::get_id()); 

        map<boost::thread::id, boost::thread*>::iterator itr; 

        itr = (m_thread_ptrs.find(this_id)); 

        if(m_thread_ptrs.end() != itr) 
        { 
            // remove it from the control of thread_group 
            m_thread_group.remove_thread(itr->second); 
            // delete it 
            delete itr->second; 

            // remove from the map 
            m_thread_ptrs.erase(this_id); 

            // log to console for debug 
            cout << "thread erased: " 
                << this_id << ", " 
                << m_thread_group.size() << ", " 
                << m_thread_ptrs.size() << "\n";             
        } 
    }               
}