Является ли getrusage нарушенным в Linux (2.6.30)

Этот код

 void print_usage(char * msg)
{
    struct rusage usage;
    getrusage(RUSAGE_SELF, &usage);
    printf("Limits: %s\n", msg);
    printf("  %s, %li\n",  " maximum resident set size "    , usage.ru_maxrss  );
    printf("  %s, %li\n",  " integral shared memory size "  , usage.ru_ixrss   );
    printf("  %s, %li\n",  " integral unshared data size "  , usage.ru_idrss   );
    printf("  %s, %li\n",  " integral unshared stack size " , usage.ru_isrss   );
    printf("  %s, %li\n",  " page reclaims "                , usage.ru_minflt  );
    printf("  %s, %li\n",  " page faults "                  , usage.ru_majflt  );
    printf("  %s, %li\n",  " swaps "                        , usage.ru_nswap   );
    printf("  %s, %li\n",  " block input operations "       , usage.ru_inblock );
    printf("  %s, %li\n",  " block output operations "      , usage.ru_oublock );
    printf("  %s, %li\n",  " messages sent "                , usage.ru_msgsnd  );
    printf("  %s, %li\n",  " messages received "            , usage.ru_msgrcv  );
    printf("  %s, %li\n",  " signals received "             , usage.ru_nsignals);
    printf("  %s, %li\n",  " voluntary context switches "   , usage.ru_nvcsw   );
    printf("  %s, %li\n",  " involuntary context switches " , usage.ru_nivcsw  );

}

выводит только нули для многих полей, даже если я использую его в довольно большой программе (после запуска jvm)

   maximum resident set size , 0
   integral shared memory size , 0
   integral unshared data size , 0
   integral unshared stack size , 0
   page reclaims , 2514
   page faults , 0
   swaps , 0
   block input operations , 0
   block output operations , 0
   messages sent , 0
   messages received , 0
   signals received , 0
   voluntary context switches , 137
   involuntary context switches , 1

Ненулевые поля: "*vcsw", "*flt".

Все *rss, *swap, msg*, *block, *signals равны нулю.

Что-то сломалось?

Linux - это x86 2.6.30.

Ответ 1