Hello,

AFAIK, POSIX specifies that pthread_cond_broadcast should wake up
waiting threads in their scheduling orders.  So, in the attached
program, main is expected to always return 1 because:

1. When main performs pthread_cond_broadcast(&b) in line 52, the
condition variable b already has threads task_1 and task_2 waiting.

2. Both task_1 and task_2 are scheduled using SCHED_FIFO such that
task_2 has a priority higher than task_1.

3. Since pthread_cond_broadcast wakes up waiting threads in their
scheduling orders, task_2 runs first and sets the value to be returned
to 2 and then task_1 runs and sets the value to be returned to 1.

4. The main function returns the last value written, which is 1.

However, I observe that the main function can return the value 2,
indicating that task_1 can run before task_2 despite POSIX
specification by the following steps:
$ gcc -O2 -o z z.c -pthread
$ sudo chown root z
$ sudo chmod u+s z
$ for ((i = 0; i < 1000000; ++i)); do ./z; if [ $? -eq 2 ]; then echo
It is two at $i; break; fi; done

When I follow the steps in my Ubuntu 16.04.6 (the output of uname -a
is: Linux Eus 4.15.0-75-generic #85~16.04.1-Ubuntu SMP Wed Jan 15
12:30:12 UTC 2020 x86_64 x86_64 x86_64 GNU/Linux), which uses Glibc
2.23 (obtained by executing /lib/x86_64-linux-gnu/libc-2.23.so -v) and
GCC 9.2.1, I get an output like the following, indicating that task_1
can run before task_2:

It is two at 22718

Could someone help me figure out why pthread_cond_broadcast does not
respect the POSIX specification, please?

Thank you.

-- 
Best regards,
Tadeus