此代码
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 );
}
只报告许多字段的零,即使我在相当大的程序中使用它(在jvm启动之后)
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
非零字段为“*vcsw
”,“*flt
”。
所有*rss
,*swap
,msg*
,*block
,*signals
均为零。
有什么东西坏了吗?
Linux是x86 2.6.30。
答案 0 :(得分:14)
是的,它部分被打破了。并非所有字段都由内核填充。 http://www.kernel.org/doc/man-pages/online/pages/man2/getrusage.2.html
工作领域:
ru_utime
ru_stime
ru_maxrss (since Linux 2.6.32)
ru_minflt
ru_majflt
ru_inblock (since Linux 2.6.22)
ru_oublock (since Linux 2.6.22)
ru_nvcsw (since Linux 2.6)
ru_nivcsw (since Linux 2.6)
未使用的字段:
ru_ixrss (unmaintained)
ru_idrss (unmaintained)
ru_isrss (unmaintained)
ru_nswap (unmaintained)
ru_msgsnd (unmaintained)
ru_msgrcv (unmaintained)
ru_nsignals (unmaintained)