c++ timer terminate without an active exception? - c++

Question

I want to design a timer in c++, to execute my function after a fixed time.
the code likes like:
#include <thread>
typedef void (*callback)();
class timer {
public:
void start(int sec, callback f) {
std::thread t([&sec, &f]() {sleep(sec); f();});
}
};
void test () {
printf("here called\n");
}
int main() {
timer t;
t.start(3, test);
while (1);
}
but when i run this code, i got:
terminate called without an active exception
[1] 3168208 abort (core dumped) ./a.out
can you help on this? and, any suggestions for a more flexible timer design?

Answer 1

You created a std::thread and destructed it without detaching it.
std::thread t([&sec, &f]() {sleep(sec);
Either call join to wait on it, or call detach.
Note also the capture by reference issue in your comments.

Answer 2

There are a few problems with your code that need to be addressed:
After spawning a std::thread you need to synchronize it using std::thread::join().
Remove the reference capture from the sec parameter in order to prevent dangling of references by the end of the scope of start().
sleep() is platform-dependent, so your code will only work for certain platforms that support it. Instead use, std::this_thread::sleep_for(std::chrono::seconds(sec));
#include <thread>
#include <chrono> // For std::chrono
#include <cstdio> // For printf
typedef void (*callback)();
class timer {
// Bring the thread object outside the function and make it an instance variable of the class
std::thread t;
public:
// Spawns a thread
void start(int const sec, callback&& f) {
if (t.joinable()) // If the object already has a thread attached to it, call 'join()' on it
t.join();
/* Capture 'sec' by value as it is a local variable, consequently, capture
'f' by reference as it is a function and its lifetime is throughout the whole
program */
t = std::thread([sec, &f]() {
std::this_thread::sleep_for(std::chrono::seconds(sec));
f();
});
}
// After the class gets destroyed, the thread is synchronized
~timer() {
t.join();
}
};
void test () {
printf("here called\n");
}
int main() {
timer t;
t.start(3, test);
}