[c++] C++ terminate called without an active exception

I am getting a C++ error with threading:

terminate called without an active exception
Aborted

Here is the code:

#include <queue>
#include <thread>
#include <mutex>
#include <condition_variable>

template<typename TYPE>
class blocking_stream
{
public:
    blocking_stream(size_t max_buffer_size_)
        :   max_buffer_size(max_buffer_size_)   
    {
    }

    //PUSH data into the buffer
    blocking_stream &operator<<(TYPE &other)
    {
        std::unique_lock<std::mutex> mtx_lock(mtx); 
        while(buffer.size()>=max_buffer_size)
            stop_if_full.wait(mtx_lock);

        buffer.push(std::move(other));

        mtx_lock.unlock();
        stop_if_empty.notify_one();
        return *this;
    }
    //POP data out of the buffer 
    blocking_stream &operator>>(TYPE &other)
    {
        std::unique_lock<std::mutex> mtx_lock(mtx);
        while(buffer.empty())
            stop_if_empty.wait(mtx_lock);

        other.swap(buffer.front()); 
        buffer.pop();

        mtx_lock.unlock();
        stop_if_full.notify_one();
        return *this;
    }

private:
    size_t max_buffer_size;
    std::queue<TYPE> buffer;
    std::mutex mtx;
    std::condition_variable stop_if_empty,
                            stop_if_full;
    bool eof;   
};

I modeled my code around this example: http://www.justsoftwaresolutions.co.uk/threading/implementing-a-thread-safe-queue-using-condition-variables.html

What am I doing wrong and how do I fix the error?

How to reproduce that error:

#include <iostream>
#include <stdlib.h>
#include <string>
#include <thread>
using namespace std;
void task1(std::string msg){
  cout << "task1 says: " << msg;
}
int main() { 
  std::thread t1(task1, "hello"); 
  return 0;
}

Compile and run:

el@defiant ~/foo4/39_threading $ g++ -o s s.cpp -pthread -std=c++11
el@defiant ~/foo4/39_threading $ ./s
terminate called without an active exception
Aborted (core dumped)

You get that error because you didn't join or detach your thread.

One way to fix it, join the thread like this:

#include <iostream>
#include <stdlib.h>
#include <string>
#include <thread>
using namespace std;
void task1(std::string msg){
  cout << "task1 says: " << msg;
}
int main() { 
  std::thread t1(task1, "hello"); 
  t1.join();
  return 0;
}

Then compile and run:

el@defiant ~/foo4/39_threading $ g++ -o s s.cpp -pthread -std=c++11
el@defiant ~/foo4/39_threading $ ./s
task1 says: hello

The other way to fix it, detach it like this:

#include <iostream>
#include <stdlib.h>
#include <string>
#include <unistd.h>
#include <thread>
using namespace std;
void task1(std::string msg){
  cout << "task1 says: " << msg;
}
int main() 
{ 
     {

        std::thread t1(task1, "hello"); 
        t1.detach();

     } //thread handle is destroyed here, as goes out of scope!

     usleep(1000000); //wait so that hello can be printed.
}

Compile and run:

el@defiant ~/foo4/39_threading $ g++ -o s s.cpp -pthread -std=c++11
el@defiant ~/foo4/39_threading $ ./s
task1 says: hello

Read up on detaching C++ threads and joining C++ threads.

As long as your program die, then without detach or join of the thread, this error will occur. Without detaching and joining the thread, you should give endless loop after creating thread.

int main(){

std::thread t(thread,1);

while(1){}

//t.detach();
return 0;}

It is also interesting that, after sleeping or looping, thread can be detach or join. Also with this way you do not get this error.

Below example also shows that, third thread can not done his job before main die. But this error can not happen also, as long as you detach somewhere in the code. Third thread sleep for 8 seconds but main will die in 5 seconds.

void thread(int n) {std::this_thread::sleep_for (std::chrono::seconds(n));}

int main() {
std::cout << "Start main\n";
std::thread t(thread,1);
std::thread t2(thread,3);
std::thread t3(thread,8);
sleep(5);

t.detach();
t2.detach();
t3.detach();
return 0;}

Eric Leschinski and Bartosz Milewski have given the answer already. Here, I will try to present it in a more beginner friendly manner.

Once a thread has been started within a scope (which itself is running on a thread), one must explicitly ensure one of the following happens before the thread goes out of scope:

• The runtime exits the scope, only after that thread finishes executing. This is achieved by joining with that thread. Note the language, it is the outer scope that joins with that thread.
• The runtime leaves the thread to run on its own. So, the program will exit the scope, whether this thread finished executing or not. This thread executes and exits by itself. This is achieved by detaching the thread. This could lead to issues, for example, if the thread refers to variables in that outer scope.

Note, by the time the thread is joined with or detached, it may have well finished executing. Still either of the two operations must be performed explicitly.

year, the thread must be join(). when the main exit

First you define a thread. And if you never call join() or detach() before calling the thread destructor, the program will abort.

As follows, calling a thread destructor without first calling join (to wait for it to finish) or detach is guarenteed to immediately call std::terminate and end the program.

Either implicitly detaching or joining a joinable() thread in its destructor could result in difficult to debug correctness (for detach) or performance (for join) bugs encountered only when an exception is raised. Thus the programmer must ensure that the destructor is never executed while the thread is still joinable.