On Tue, Nov 10, 2020 at 11:12 PM Andrey Konovalov <andreyk...@google.com> wrote:
>
> From: Vincenzo Frascino <vincenzo.frasc...@arm.com>
>
> This test is specific to MTE and verifies that the GCR_EL1 register
> is context switched correctly.
>
> It spawn 1024 processes and each process spawns 5 threads. Each thread

Nit: "spawns"


> +       srand(time(NULL) ^ (pid << 16) ^ (tid << 16));
> +
> +       prctl_tag_mask = rand() % 0xffff;

Nit: if you want values between 0 and 0xffff you probably want to use
bitwise AND.


> +
> +int execute_test(pid_t pid)
> +{
> +       pthread_t thread_id[MAX_THREADS];
> +       int thread_data[MAX_THREADS];
> +
> +       for (int i = 0; i < MAX_THREADS; i++)
> +               pthread_create(&thread_id[i], NULL,
> +                              execute_thread, (void *)&pid);

It might be simpler to call getpid() in execute_thread() instead.

> +int mte_gcr_fork_test()
> +{
> +       pid_t pid[NUM_ITERATIONS];
> +       int results[NUM_ITERATIONS];
> +       pid_t cpid;
> +       int res;
> +
> +       for (int i = 0; i < NUM_ITERATIONS; i++) {
> +               pid[i] = fork();
> +
> +               if (pid[i] == 0) {

pid[i] isn't used anywhere else. Did you want to keep the pids to
ensure that all children finished the work?
If not, we can probably go with a scalar here.


> +       for (int i = 0; i < NUM_ITERATIONS; i++) {
> +               wait(&res);
> +
> +               if(WIFEXITED(res))
> +                       results[i] = WEXITSTATUS(res);
> +               else
> +                       --i;

Won't we get stuck in this loop if fork() returns -1 for one of the processes?

> +       }
> +
> +       for (int i = 0; i < NUM_ITERATIONS; i++)
> +               if (results[i] == KSFT_FAIL)
> +                       return KSFT_FAIL;
> +
> +       return KSFT_PASS;
> +}
> +


-- 
Alexander Potapenko
Software Engineer

Google Germany GmbH
Erika-Mann-Straße, 33
80636 München

Geschäftsführer: Paul Manicle, Halimah DeLaine Prado
Registergericht und -nummer: Hamburg, HRB 86891
Sitz der Gesellschaft: Hamburg

Reply via email to