Reputation: 17493
I would like to measure how long a particular operation takes, therefore I've written the following piece of code:
for (int counter2 = 0; counter2 <= 10; counter2++) { // I'll do the test 10 times,
// for having reliable results
time_t ltime;
time(<ime); // What's the time at the beginning?
for (counter = 0; counter <= 1000000; counter++) {
Do_Something();
}
time_t ltime2;
time(<ime2); // What's the time after having launched Do_Something()?
printf("Times:First=[%d], Second=[%d]\n", ltime, ltime2);
}
I'm expecting something like:
Times: First=[1559574983], Second=[1559574985]
Times: First=[1559574990], Second=[1559574999]
Instead I get this:
Times: First=[1559574983], Second=[0]
Times: First=[1559574990], Second=[0]
I've already debugged, and ltime2
seems to be correct. What am I doing wrong?
Upvotes: 0
Views: 117
Reputation: 17493
Thanks for the quick replies.
Just for the record, hereby the final solution I'm using (combining the mentioned other Stackoverflow post and difftime()
function):
printf("First=[%ju], Second=[%ju], Difference=[%f]\n", (uintmax_t) ltime,
(uintmax_t) ltime2,
difftime(ltime2, ltime));
Upvotes: 0
Reputation: 32586
The format "%d"
is not the right one in your case, but you don't have to take care of the type supporting time_t
just replace
printf("Times:First=[%d], Second=[%d]\n", ltime, ltime2);
by
std::cout << "Times:First=[" << ltime << "], Second=[" << ltime2 << "]" << std::endl;
If for an obscure reason you really want to use a C printf as proposed by @Lightness Races in Orbit 3 in a remark you can convert the value to an other type (hopping enough long) and use the format compatible to that type, see Format specifier to print time(0) in C
Upvotes: 4