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linux@64 获取时间的性能评估
听人说gettimeofday 在64bit下有缓存，速度很快，测试下了，感觉不对啊。。
#include <time.h>#include <sys/time.h>#include <stdio.h>#include <stdint.h>int foo(int i){ return i;}const int64_t MAX_COUNT = 100000*1000;struct TimerEval { TimerEval(const char* module) { start_time_ = time(NULL); module_ = module; } ~TimerEval() { time_t end_time = time(NULL); printf("%s\telapse : %d sec\n", module_, (end_time - start_time_)); } time_t start_time_; const char* module_;};int main(){ struct timeval tpTmp; printf("repeat %d times, test result is : \n", MAX_COUNT); { TimerEval eval("call fun"); for (int i=0; i<MAX_COUNT; ++i) foo(i); } { TimerEval eval("call time"); for (int i=0; i<MAX_COUNT; ++i) time(NULL);; } { TimerEval eval("call gettimeofday"); for (int i=0; i<MAX_COUNT; ++i) gettimeofday(&tpTmp, NULL);; } { TimerEval eval("call clock_gettime"); struct timespec tp; for (int i=0; i<MAX_COUNT; ++i) clock_gettime(CLOCK_REALTIME, &tp); } return 0;}
测试结果
repeat 100000000 times, test result is :
call fun elapse : 1 sec
call time elapse : 1 sec
call gettimeofday elapse : 7 sec
call clock_gettime elapse : 15 sec
编译参数
g++ timer_benchmarck.cc -m64 -lrt
貌似事实可能不是这样，求教于大家，可能是什么原因。
如果说time只是在gettimeofday的基础上封装了一层，那怎么time会比gettimeofday还快，不科学啊！
/* Return the current time as a `time_t‘ and also put it in *T if T is not NULL. Time is represented as seconds from Jan 1 00:00:00 1970. */time_ttime (t) time_t *t; { struct timeval tv; time_t result; if (__gettimeofday (&tv, (struct timezone *) NULL)) result = (time_t) -1; else result = (time_t) tv.tv_sec; if (t != NULL) *t = result; return result;}
linux@64 获取时间的性能评估
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