struct timeval first, second, lapsed;
struct timezone tzp;
gettimeofday (&first, &tzp);
sleep(1);
gettimeofday (&second, &tzp);
lapsed.tv_usec = second.tv_usec - first.tv_usec;
lapsed.tv_sec = second.tv_sec - first.tv_sec;
printf("%d %d\n", lapsed.tv_sec, lapsed.tv_usec);
gettimeofday()
Author
I'm trying to use gettimeofday in linux but I don't really understand how it works. I wrote a little piece of code to check it out, but I'm still confused. Here's the code:
I'm always getting 1, when printing the seconds passed but I'm not getting a constant number when printing the microseconds, why is that?
[Edited by - Deliverance on August 12, 2007 6:16:15 AM]
Here's my little benchmark template
static inline void dosomething() {}int main( int argc, char** args ) { struct timeval t1,t2; double time1, time2; gettimeofday(&t1, NULL); dosomething(); gettimeofday(&t2, NULL); time1 = t1.tv_usec; time1 = time1/1000000.0; time1 = time1 + t1.tv_sec; time2 = t2.tv_usec; time2 = time2/1000000.0; time2 = time2 + t2.tv_sec; cout << fixed << time2-time1 << "seconds" << endl; return 0;}Quote:Original post by Deliverance
I'm always getting 1, when printing the seconds passed but I'm not getting a constant number when printing the microseconds, why is that?
Because depending on the load and granularity of the operating system scheduler, sleep(n) presents only a minimum elapsed time - in reality, it will vary, and offers no realtime guarantees. The time between sleep and wake up can be expected to be different each time that code is run.
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