CODEWITHSUNDEEP
c++c++11stdthreadtemporary-objectsstdmutex
rsy56640
rsy56640

Reputation: 309

Why is a temporary std::lock_guard object immediately unlocked?

I'm learning about std::mutex, std::thread and I am surprised at the different behavior of 2 pieces of code below:

#include <iostream>
#include <mutex>
#include <thread>

std::mutex mtx;

void foo(int k)
{
    std::lock_guard<std::mutex> lg{ mtx };
    for (int i = 0; i < 10; ++i)
        std::cout << "This is a test!" << i << std::endl;
    std::cout << "The test " << k << " has been finished." << std::endl;
}

int main()
{
    std::thread t1(foo, 1);
    std::thread t2(foo, 2);
    t1.join();
    t2.join();
    return 0;
}

The output is sequential. But if I donot name variable std::lock_guard<std::mutex>, the output is unordered

void foo(int k)
{
    std::lock_guard<std::mutex> { mtx }; // just erase the name of variable
    for (int i = 0; i < 10; ++i)
        std::cout << "This is a test!" << i << std::endl;
    std::cout << "The test " << k << " has been finished." << std::endl;
}

It seems like std::lock_guard is no use in 2nd case, Why?

Upvotes: 7

Views: 1497

Answers (1)

lubgr
lubgr

Reputation: 38315

This declaration

std::lock_guard<std::mutex> { mtx };

doesn't bind the created object to a name, it's a temporary variable that exists only for this particular statement. Opposed to that, a variable that has a name and is created on the stack lives until the end of the scope in which it's created.

In this CppCon talk (starting at 31:42), the presenter lists the creation of temporary std::lock_guard instances not bound to a local variable as a common bug in the Facebook code base.

Upvotes: 18

Related Questions