您好我正在尝试获取程序消耗的系统时间和用户时间。我遇到了一些麻烦,因为我的时间总是零毫秒。我想我做错了什么。你能指导我正确的方向吗?我的两个方法都返回0
typedef struct{
struct timeval ru_utime; /* user time used */
struct timeval ru_stime; /* system time used */
long ru_maxrss; /* maximum resident set size */
long ru_ixrss; /* integral shared memory size */
long ru_idrss; /* integral unshared data size */
long ru_isrss; /* integral unshared stack size */
long ru_minflt; /* page reclaims */
long ru_majflt; /* page faults */
long ru_nswap; /* swaps */
long ru_inblock; /* block input operations */
long ru_oublock; /* block output operations */
long ru_msgsnd; /* messages sent */
long ru_msgrcv; /* messages received */
long ru_nsignals; /* signals received */
long ru_nvcsw; /* voluntary context switches */
long ru_nivcsw; /* involuntary context switches */
}rusage;
/*************************************************************/
double getsystemtime(){
struct timeval tim;
struct rusage ru;
getrusage(RUSAGE_SELF, &ru);
tim = ru.ru_stime;
double t = (double)tim.tv_sec / 1000.0;
return t;
}
double getusertime(){
struct timeval tim;
struct rusage ru;
getrusage(RUSAGE_SELF, &ru);
tim = ru.ru_utime;
double t = (double)tim.tv_sec / 1000.0;
return t;
}
答案 0 :(得分:4)
不应该是tim.tv_sec * 1000.0 + tim.tv_usec / 1000.0转换为毫秒?您将tim.tv_sec除以1000,因此如果您的运行时间不到1000秒,您将获得0的返回值。
答案 1 :(得分:0)
在每个功能中,更改为:
double t = (double)tim.tv_sec*1000.0 + tim.tv_usec/1000.0;
不要忘记包含sys/resource.h