`time` benchmarking utility: What does `user`+`sys` = 0 actually mean? - c++

Question

One comparison I felt that was missing from the Why is this C++ program so incredibly fast? discussion is Fortran. I translated Sven Hager's C++ benchmark:
#include <iostream>
#include <cstdlib>
#include <cstdint>
int main(int argc, char* argv[]) {
uint32_t s = 0;
uint32_t outer = atoi(argv[1]);
uint32_t inner = atoi(argv[2]);
for (uint32_t i = 0; i < outer; ++i) {
for (uint32_t j = 0; j < inner; ++j)
++s;
s -= inner;
}
std::cout << s << std::endl;
return 0;
}
to its Fortran equivalent:
PROGRAM Benchmark
IMPLICIT NONE
INTEGER :: i,j,s
INTEGER, PARAMETER :: outer=1000,inner=1000000
s = 0
DO i = 1, outer
DO j = 1, inner
s = s + 1
END DO
s = s - inner
END DO
PRINT *, s
END PROGRAM Benchmark
and compiled a fully optimized version with gfortran -g -std=f2008 -Wall -Wextra -O3 Benchmark.f08. I expected to acheive similar performance as Herr Hager:
real 0m0.003s
user 0m0.002s
sys 0m0.002s
What I got was a little puzzling:
real 0m0.003s
user 0m0.000s
sys 0m0.000s
Digging more deeply, I found this discussion on What do 'real', 'user' and 'sys' mean in the output of time(1)?. In it they say that user+sys gives the actual CPU time the process used. So, what does a user+sys of zero actually mean?

Answer 1

In this case, "zero" means "less than a millisecond" (or half a millisecond, depending on how it's rounded), since that's the resolution of the times given by time.
It's only useful for measuring programs that take considerably longer than a millisecond to run.
The reason that the Fortran version is much faster is probably because the loop bounds are hard-coded constants, so that the entire calculation can be done at compile time, leaving just PRINT 0 to do at runtime.