能否请您查看并提出此代码有什么问题? 它要么在第21行崩溃(cond_var_.wait(lock);在gc_thread_proc()中),要么在第56行锁定(lock.lock(); release()中)。
#include <condition_variable>
#include <deque>
#include <functional>
#include <mutex>
#include <thread>
#include <vector>
#include <iostream>
class stream {
std::deque<int> pending_cleanups_;
std::mutex mut_{};
bool continue_{true};
std::thread gc_worker_;
std::condition_variable cond_var_;
void gc_thread_proc() {
while (true) {
std::vector<int> events_to_clean;
std::unique_lock<std::mutex> lock(mut_);
while (pending_cleanups_.empty() && continue_) {
cond_var_.wait(lock);
}
if (!continue_) {
break;
}
std::move(std::begin(pending_cleanups_), std::end(pending_cleanups_), std::back_inserter(events_to_clean));
pending_cleanups_.clear();
}
}
public:
explicit stream() : gc_worker_(&stream::gc_thread_proc, this) {}
void register_pending_event(int val) {
{
std::lock_guard<std::mutex> lock_guard(mut_);
pending_cleanups_.push_back(val);
}
cond_var_.notify_one();
}
void release() {
std::unique_lock<std::mutex> lock(mut_);
if (!continue_) {
return;
}
continue_ = false;
lock.unlock();
cond_var_.notify_one();
gc_worker_.join();
lock.lock();
pending_cleanups_.clear();
}
~stream() { release(); }
};
int main() {
int N=100000;
while(N--) {
std::cout << ".";
stream s;
}
std::cout << "ok";
return 0;
}
更改成员顺序可以解决此问题-将cond_var_放在gc_worker_问题不会重现之前。但是我想它并不能解决问题,只是以某种方式将其隐藏了...
答案 0 :(得分:3)
非静态数据成员按照类定义中的声明顺序初始化:https://en.cppreference.com/w/cpp/language/initializer_list
3) Then, non-static data members are initialized in order of declaration in the class definition.
在您的情况下,由于已将std :: thread成员初始化为开始在其构造函数中执行,因此在gc_thread_proc中使用cv时可能不会初始化cv。拥有std :: thread成员的一种命令方式是将其分配给类构造函数,即
class stream {
std::thread gc_worker_;
std::condition_variable cond_var_;
public:
stream(): {
gc_work = std::move(std::thread(&stream::gc_thread_proc, this));
}
};