#include <iostream>
#include <thread>
#include <chrono>
#include <mutex>
#include <thread>
std::mutex mtx;
void func2() {
mtx.lock();
std::cout << "here is 2" << std::endl;
mtx.unlock();
}
void func1() {
mtx.lock();
std::cout << "here is 1" << std::endl;
func2();
mtx.unlock();
}
int main() {
func1();
}
but if I modify the main func as follows, it cause dead lock
int main() {
std::thread t1(func1);
t1.join();
}
I complied both two by "g++ test.cpp -std=c++11 -lpthread"
Calling lock in the same thread twice (without unlocking the mutex) is undefined. From cppreference:
If lock is called by a thread that already owns the mutex, the behavior is undefined: for example, the program may deadlock.
It may deadlock. Or it may not. The behavior is undefined.
Note that std::recursive_mutex can be locked multiple times (though only up to some unspecified limit). However, code that needs a recursive mutex is more complicated. In your example it would be easier to remove the locking from func1, because it is only called when the mutex is already locked. In general it isn't that simple.
Test environment: Ubuntu 18.04.3 LTS g++ (Ubuntu 7.5.0-3ubuntu1~18.04) 7.5.0.
Can std::mutex be reentrant? Why does the following test code 1 pass?
code1:
#include <iostream>
#include <mutex>
std::mutex g_mtx4val;
int g_val = 5;
void test() {
std::lock_guard<std::mutex> lck(g_mtx4val);
std::cout << "g_val=" << g_val << std::endl;
if (g_val > 0) {
--g_val;
test();
}
}
int main() {
test();
std::cout << "done ...." << std::endl;
return 0;
}
peanut#peanut:~/demo$ g++ test.cpp
peanut#peanut:~/demo$ ./a.out
g_val=5
g_val=4
g_val=3
g_val=2
g_val=1
g_val=0
done ...
code2:
// Same code 1
int main() {
std::thread t1(test);
t1.join();
std::cout << "done ...." << std::endl;
return 0;
}
peanut#peanut:~/demo$ g++ test2.cpp -lpthread
peanut#peanut:~/demo$ ./a.out
g_val=5
^C
peanut#peanut:~/demo$
code2 has a deadlock.
Why code1 can pass the test?
From the documentation page:
mutex offers exclusive, non-recursive ownership semantics
So the answer to the question in the title is no.
Can std::mutex be reentrant?
No, but if you want a recursive mutex, the std::recursive_mutex class provides that functionality.
Why does the following test code 1 pass?
What behavior were you expecting to see? The std::mutex documentation page simply says:
A calling thread must not own the mutex prior to calling lock or try_lock.
... it doesn't say what will happen if the calling thread breaks the above rule; which means that a program that breaks the rule may "appear to work", but even so is still incorrect and buggy.
As we know, using boehm-gc in multi-thread requires calling GC_register_my_thread with stack base from GC_get_stack_base. but It seems not to work well with C++11's thread library, such as std::thread... How can I use boehm-gc with C++11's thread library?
(I use VS2013)
edit: This is tested code. std::thread is good, but std::future doesn't work (stop on _CrtIsValidHeapPointer
#include <iostream>
#include <thread>
#include <future>
#define GC_THREADS
#include <gc.h>
#include <gc_cpp.h>
#pragma comment(lib, "gcmt-lib")
void foo()
{
GC_stack_base sb;
GC_get_stack_base(&sb);
GC_register_my_thread(&sb);
int *ptr;
for (int i = 0; i < 10; i++)
{
ptr = new (GC) int;
*ptr = 1;
}
GC_unregister_my_thread();
}
int main()
{
GC_INIT();
GC_allow_register_threads();
std::cout << "test for std::thread";
std::thread thrd(foo);
thrd.join();
std::cout << " [sucs]\n";
std::cout << "test for std::future";
std::future<void> fu = std::async(std::launch::async, foo);
fu.get();
std::cout << " [sucs]\n";
std::cin.get();
}
edit: here is a capture of stack trace (Sorry that it isn't English, but I think it doesn't matter, anyway)
and here is a debug message
HEAP[TestGC.exe]: Invalid address specified to RtlValidateHeap( 00E80000, 00C92F80 )
While debugging, I found The error occurs after fu.get().
edit: The error doesn't occur with /MD(or /MDd)...
(I think GC might touch library's pointers (namespcae Concurrency), but it is just guess;;)
Before you start using the collector and before you create the threads make sure that you issue both
GC_INIT, and
GC_allow_register_threads
Then in every thread follow it up with,
GC_get_stack_base/GC_register_my_thread, and eventually
GC_unregister_my_thread.
You didn't say what you are compiling with but it works for gcc 4.8 (with -std=c++11).
EDIT: The OP was able to resolve the issue by addressing the instruction above and compiling the code with the /MD[d] flags for the multi-threaded dynamic MSVCR100 runtime. The issue remained unresolved for the multithreaded statically compiled runtime.
Take the following code:
#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <chrono>
using namespace std;
int main() {
mutex m;
condition_variable c;
bool fired = false;
int i = 0;
// This thread counts the times the condition_variable woke up.
// If no spurious wakeups occur it should be close to 5.
thread t([&]() {
unique_lock<mutex> l(m);
while (!fired) {
c.wait_for(l, chrono::milliseconds(100));
++i;
}
});
// Here we wait for 500ms, then signal the other thread to stop
this_thread::sleep_for(chrono::milliseconds(500));
{
unique_lock<mutex> l(m);
fired = true;
c.notify_all();
cout << i << endl;
}
t.join();
}
Now, when I build this using clang++ -std=c++11 -pthread foo.cpp everything is fine, it outputs 4 on my machine. When I build it with g++ -std=c++11 -pthread foo.cpp however I get something very large every time, e.g. 81513. I realize the number of spurious wakeups is undefined, but I was surprised to see it so high.
Additional information: When I replace the wait_for by a simple wait both clang and g++ output 0.
Is this a bug / feature in g++? Why is it even different from clang? Can I get it to behave more reasonably?
Also: gcc version 4.7.3 (Debian 4.7.3-4).
I managed to get g++-4.8 running, and the problem is gone. Very weird, seems like a bug in g++-4.7.3, although I wasn't able to reproduce it on another machine.
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?
When a thread object goes out of scope and it is in joinable state, the program is terminated. The Standard Committee had two other options for the destructor of a joinable thread. It could quietly join -- but join might never return if the thread is stuck. Or it could detach the thread (a detached thread is not joinable). However, detached threads are very tricky, since they might survive till the end of the program and mess up the release of resources. So if you don't want to terminate your program, make sure you join (or detach) every thread.
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.
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.
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.
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;}
yes, the thread must be join(). when the main exit