What is the unit of gettimeofday()?

Question

I've a program to calculate the latency of an object in a pub-sub model. I've used the following function for timestamp:

uint64_t GetTimeStamp() {
    struct timeval tv;
    gettimeofday(&tv,NULL);
    return tv.tv_sec*(uint64_t)1000000+tv.tv_usec;
}

The latency is measured as timestamp difference in publisher and subscriber. So, I'm concerned about the unit of the latency measured. Is it in seconds or microseconds??

Vocation

man gettimeofday to see struct timeval details – Mong

Just simply look with what constant you are multiplying seconds to get something else. 1sec = 1000000 microseconds. – Endosteum

pubs.opengroup.org/onlinepubs/009695399/basedefs/sys/… :) – Bradley

Strictly speaking, that's the unit of reporting, not measurement. – Cypro

If you have C++11 (gcc 4.7.2+), look at std::chrono (cppreference has a good example), for an example usage with milliseconds: daniweb.com/software-development/cpp/code/445750/…. One of the nice things about std::chrono is a quick search and replace on "milliseconds" -> "microseconds" or "nanoseconds" and you have that resolution. – Extrude

Answer 1

The timeval structure has tv_sec, which gives you the absolute value for the seconds, and tv_usec, which gives you the remaining fraction in micro seconds.

So, you could get the resolution in micro seconds.

For more information, http://www.gnu.org/software/libc/manual/html_node/Elapsed-Time.html

Puttyroot

Answer 2

tv.tv_sec gives the second count, tv.tv_usec gives the remaining microsecond count.

For gettimeofday() principle and its accuracy:

How is the microsecond time of linux gettimeofday() obtained and what is its accuracy?

Is gettimeofday() guaranteed to be of microsecond resolution?

Schilling