std :: thread和std :: endl没有预期的输出

Question

I am trying out some examples from C++11 threads I see some surprising results. With the following code

#include <iostream>
#include <thread>

void hello() {
    std::cout << "Hello concurrent world " << std::endl;
}
void do_something() {
    std::cout << "[background_task] --> [ do_something ]" << std::endl;
}
void do_something_else() {
    std::cout << "[background_task] --> [ do_something_else ]" << std::endl;
}
class background_task {
public:
    void operator()() const     {       
        do_something();
        do_something_else();
    }
};

int main ( int argc, char **argv) {
    std::thread t(hello);
    background_task bt;
    std::thread fn_obj_th(bt);
    t.join();
    fn_obj_th.join();
}

output is as follows

Hello concurrent world [background_task] --> [ do_something ]

[background_task] --> [ do_something_else ]
Press any key to continue . . .

If I replace std::cout << "Hello concurrent world " << std::endl; with std::cout << "Hello concurrent world \\n";

Result is

Hello concurrent world
[background_task] --> [ do_something ]
[background_task] --> [ do_something_else ]

Why in case with std::endl I am not getting expected output.

Answer 1

This:

std::cout << "Hello concurrent world " << std::endl;

is two separate outputs. While std::cout is thread-safe, that doesn't mean that two separate invocations of it is guaranteed to be atomic. A single output is atomic, but not two.

If you want a particular expression to be guaranteed to be atomically output, then you need to add your own synchronization primitives on top of std::cout .