Why this lambda function in loop didn't capture the parameter by value?

Question

I have following block of code, what I expected is the i inside the lambda capture list is passed by value and the i be outputted should be different.

However, the actual result is the output 20 lines of 19 is completed.

I tried to change the this line std::thread t([&func](){ to std::thread t([func](){, the output could print different i value.

My question is why std::thread t([&func](){ and std::thread t([func](){ will lead the different output value of i?

void DoSomething(std::function<void()> func)
{
    std::thread t([&func](){
        //do something
        sleep(1);
        if (func)
            func();
    });
    t.detach();
}

int main(int argc, const char * argv[]) {
    std::mutex mtx;
    for (int i = 1 ; i < 20; i ++) {
        DoSomething([i, &mtx](){
            std::unique_lock<std::mutex> lock(mtx);
            std::cout << i << " is completed" << std::endl;;
        });
    }
    sleep(10);
}

molbdnilo · Accepted Answer · 2016-05-07 08:25:16Z

2

You're capturing a reference to DoSomething's parameter, func.
By the time your thread executes, that parameter's lifetime has ended so you have a dangling reference and using it is undefined.

answered May 7, 2016 at 8:25

molbdnilo

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2 Comments

2power10 Over a year ago

is that means the func passed to the DoSomething carry the information about the capture list?

molbdnilo Over a year ago

@2power10 I'm not sure what you're asking. If you're capturing by reference you need to ensure that the referenced object is valid when you use it, otherwise the program is undefined. Capturing by value is the correct thing to do here.

MNS · Accepted Answer · 2016-05-07 08:48:40Z

0

Thread object t in DoSomething() receives a reference to the func object which is passed as by value. When func() is executed, it might be waiting for the mutex by the call std::unique_lock<std::mutex> lock(mtx); In the meantime t.detach() will be invoked and the function would have exited by destroying the by value parameter func. This will result in dangling reference as said by @molbidnilo.

The problem can be solved by calling t.join(); instead of t.detach().

answered May 7, 2016 at 8:48

MNS

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1 Comment

2power10 Over a year ago

use t.join() will block the DoSomething, and that's why I use t.detach() here. I think the problem can be solved by capturing the func by value instead of reference.

Tony · Accepted Answer · 2016-05-07 10:29:40Z

0

if you use t.detach(), the thread is running independently, now that you pass function object by reference, so if the function object was destroyed when the independent thread is trying to refer, wouldn't that cause crash of the program. but if you pass the function object by value, this issue is solved. i tested your code in my machine, it crashed when using passed function object by reference.

answered May 7, 2016 at 10:29

Tony

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