Different behavior when `std::lock_guard<std::mutex>` object has no name

问题

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>
using namespace std;

std::mutex mtx;

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

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

回答1:

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.

来源：https://stackoverflow.com/questions/51963131/different-behavior-when-stdlock-guardstdmutex-object-has-no-name