C++ multithread program I/O problem with std::endl - c++

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. E.g.,
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.