Why does my thread not run in background?

Question

In the listing bellow, I expect that as I call t.detach() right after the line when the thread is created, the thread t will run in background while the printf("quit the main function now \n") will called and thenmain will exit.

#include <thread>
#include <iostream>

void hello3(int* i)
{

    for (int j = 0; j < 100; j++)
    {
        *i = *i + 1;
        printf("From new thread %d \n", *i);
        fflush(stdout);

    }

    char c = getchar();
 }

int main()
{
    int i;
    i = 0;
    std::thread t(hello3, &i);
    t.detach();
    printf("quit the main function now \n");
    fflush(stdout);
    return 0;
}

However from what it prints out on the screen it is not the case. It prints

From new thread 1
From new thread 2
....
From new thread 99
quit the main function now.

It looks like the main function waits until the thread finishes before it executes the commandprintf("quit the main function now \n"); and exits.

Can you please explain why it is? What I am missing here?

Answer 1

The problem is that your thread is too fast.

It's able to print all 100 strings before main gets a chance to continue.

Try making the thread slower and you'll see the main printf before that of the thread.

Answer 2

It happens, based on your OS scheduling. Moreover the speed of your thread affects the output too. If you stall the thread (change 100 to 500 for example), you will see the message first.

I just executed the code and the "quit the main function now." message appeared first, like this:

quit the main function now 
From new thread 1 
From new thread 2 
From new thread 3 
From new thread 4 
...

You are right about detach:

Detaches the thread represented by the object from the calling thread, allowing them to execute independently from each other.

but this does not guarantee that the message "quit the main function now" will appear first, although it's very likely.