C ++多线程程序I / O问题与std :: endl

Question

I try to learn how to use C++11 thread library and then, I am confused about the output of my following code.

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

std::mutex mtx;

void thread_function()
{
    std::cout << "Inside Thread :: ID = " << std::this_thread::get_id() << std::endl;
}

int main()
{
    std::thread threadObj1(thread_function);
    std::thread threadObj2(thread_function);

    if (threadObj1.get_id() != threadObj2.get_id())
        std::cout << "Both threads have different id" << std::endl;

    threadObj1.join();
    threadObj2.join();
    std::cout<<"From Main Thread :: ID of Thread 1 = "<<threadObj1.get_id()<<std::endl;
    std::cout<<"From Main Thread :: ID of Thread 2 = "<<threadObj2.get_id()<<std::endl;


    return 0;
}

I attach every std::cout with std::endl in order to flush the buffer and output the '/n' character. However, finally I got the output as the following.

Both threads have different idInside Thread :: ID = Inside Thread :: ID = 
0x700003715000
0x700003692000
From Main Thread :: ID of Thread 1 = 0x0
From Main Thread :: ID of Thread 2 = 0x0
Program ended with exit code: 0

It seems that the '/n' before the Inside Thread disappeared. Could you please help me figure it out? Thank you so much!

Answer 1

You have 3 threads which are accessing cout without any synchronization. You have defined mtx but it is not used, why?

Add lock_guard to protect cout statement:

void thread_function()
{
    std::lock_guard<std::mutex> lk{mtx};
    std::cout << "Inside Thread :: ID = " << std::this_thread::get_id() << std::endl;
}

    if (threadObj1.get_id() != threadObj2.get_id())
    {
        std::lock_guard<std::mutex> lk{mtx};
        std::cout << "Both threads have different id" << std::endl;
    }

Answer 2

I think I should also receive three '\\n' right?

The three '\\n' characters in question are there in your output. They're at the ends of the first three lines of output.

I think maybe you misunderstand what this line from your example means:

std::cout << "Inside Thread :: ID = " << std::this_thread::get_id() << std::endl;

There are four separate function calls explicit in that one line of code. That one line does exactly the same thing as these four lines:

std::cout << "Inside Thread :: ID = ";
auto id = std::this_thread::get_id();
std::cout << id;
std::cout << std::endl;

Even assuming that the std::cout object is fully synchronized, You have done nothing to prevent the various threads from interleaving the separate function calls. Eg,

• main thread calls std::cout << "Both threads have different id";
• threadObj1 calls std::cout << "Inside Thread :: ID = ";
• threadObj2 calls std::cout << "Inside Thread :: ID = ";
• main thread calls std::cout << std::endl;
• threadObj1 calls std::cout << std::this_thread::get_id();
• threadObj1 calls stc::cout << std::endl;
• etc.