How to implement timeout for function in c++

Solution 1:

You can create a separate thread to run the call itself, and wait on a condition variable back in your main thread which will be signalled by the thread doing the call to f once it returns. The trick is to wait on the condition variable with your 1s timeout, so that if the call takes longer than the timeout you will still wake up, know about it, and be able to throw the exception - all in the main thread. Here is the code (live demo here):

#include <iostream>
#include <chrono>
#include <thread>
#include <mutex>
#include <condition_variable>

using namespace std::chrono_literals;

int f()
{
    std::this_thread::sleep_for(10s); //change value here to less than 1 second to see Success
    return 1;
}

int f_wrapper()
{
    std::mutex m;
    std::condition_variable cv;
    int retValue;

    std::thread t([&cv, &retValue]() 
    {
        retValue = f();
        cv.notify_one();
    });

    t.detach();

    {
        std::unique_lock<std::mutex> l(m);
        if(cv.wait_for(l, 1s) == std::cv_status::timeout) 
            throw std::runtime_error("Timeout");
    }

    return retValue;    
}

int main()
{
    bool timedout = false;
    try {
        f_wrapper();
    }
    catch(std::runtime_error& e) {
        std::cout << e.what() << std::endl;
        timedout = true;
    }

    if(!timedout)
        std::cout << "Success" << std::endl;

    return 0;
}

Solution 2:

You can also use std::packaged_task to run your function f() in another thread. This solution is more or less similar to this one, only that it uses standard classes to wrap things up.

std::packaged_task<void()> task(f);
auto future = task.get_future();
std::thread thr(std::move(task));
if (future.wait_for(1s) != std::future_status::timeout)
{
   thr.join();
   future.get(); // this will propagate exception from f() if any
}
else
{
   thr.detach(); // we leave the thread still running
   throw std::runtime_error("Timeout");
}

You can probably even try to wrap it into a function template, to allow calling arbitrary functions with timeout. Something along the lines of:

template <typename TF, typename TDuration, class... TArgs>
std::result_of_t<TF&&(TArgs&&...)> run_with_timeout(TF&& f, TDuration timeout, TArgs&&... args)
{
    using R = std::result_of_t<TF&&(TArgs&&...)>;
    std::packaged_task<R(TArgs...)> task(f);
    auto future = task.get_future();
    std::thread thr(std::move(task), std::forward<TArgs>(args)...);
    if (future.wait_for(timeout) != std::future_status::timeout)
    {
       thr.join();
       return future.get(); // this will propagate exception from f() if any
    }
    else
    {
       thr.detach(); // we leave the thread still running
       throw std::runtime_error("Timeout");
    }
}

And then use:

void f1() { ... }
call_with_timeout(f1, 5s);

void f2(int) { ... }
call_with_timeout(f2, 5s, 42);

int f3() { ... }
int result = call_with_timeout(f3, 5s);

This is an online example: http://cpp.sh/7jthw

Solution 3:

You can create a new thread and asynchronously wait for 1s to pass, and then throw an exception. However, exceptions can only be caught in the same thread where they're thrown, so, you cannot catch in the same thread where you called f(), like in your example code - but that's not a stated requirement, so it may be OK for you.

Only if f is guaranteed to return in less than 1s, can you do this synchronously:

• store current time
• call f()
• wait for current time - stored time + 1s

But it may be quite difficult to prove that f in fact does return in time.