How to use c time in linux to print the function running time?

Question

when I run c code in linux,the code always doesnt print out the elapse time,and the result always is 0.The code is just as follow:

#include <sys/time.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
void main(int argc,char* argv[]){
  int n;
  if(argc == 2){
    n = atoi(argv[1]);
  }
  struct timeval start, end;
  gettimeofday(&start, 0);
  int r = fib(n);
  gettimeofday(&end, 0);
  long mtime, s,us;
  s = end.tv_sec  - start.tv_sec;
  us = end.tv_usec - start.tv_usec;
  printf("s=%f,us=%f  \n", s, us);
  mtime = (s*1000 + us/1000.0)+0.5;
  printf("Fib result for %d is: %d;elapsing %f \n", n, r, mtime); 

}

int fib(int n){
  if(n == 0) return 0;
  if(n == 1) return 1;
  return fib(n-1)+fib(n-2);
}

score 6 · Answer 1 · answered Nov 23 '11 at 09:45

Don't overlook your compiler warnings; you're trying to print three long variables (mtime, s, and us) as if they were doubles:

fib.c: In function ‘main’:
fib.c:17:3: warning: format ‘%f’ expects type ‘double’, but argument 2 has type ‘long int’
fib.c:17:3: warning: format ‘%f’ expects type ‘double’, but argument 3 has type ‘long int’
fib.c:19:3: warning: format ‘%f’ expects type ‘double’, but argument 4 has type ‘long int’

Change s and us to long, and change the format for s and us to %ld, and the program compiles (and runs) without fault.

Basile Starynkevitch · Accepted Answer · 2013-09-13T05:20:14.580

All the suggestions do in fact work, but the granularity of the time measurement is big (typically 10 to 100 milliseconds). So it actually measure something for a computation which last e.g. half a second. On current processors (running at 2 to 3Ghz, with about 3-5 instructions per cycle), that means something like a billion machine instructions executed (an "elementary step" in our C program -with an ill-defined notion of step is usually a dozen machine instructions). So your test is too small, you really should compute a million times fibionacci (10).

To be more specific the program below (where some computations are output, to avoid optimizing them all) is running in about 2 seconds. (on million computations of fibionacci of something less than 16).

#include <stdio.h>
#include <unistd.h>
#include <time.h>
long fib(int n){
  if(n == 0) return 0;
  if(n == 1) return 1;
  return fib(n-1)+fib(n-2);
}

int main ()
{
  int i=0;
  int p = (int) getpid();
  clock_t cstart = clock();
  clock_t cend = 0;
  for (i=0; i<1000000; i++) {
    long f = fib(i%16);
    if (i % p == 0) printf("i=%d, f=%ld\n", i, f);
  }
  cend = clock();
  printf ("%.3f cpu sec\n", ((double)cend - (double)cstart)* 1.0e-6);
  return 0;
}

The last few lines output with time ./fib (compiled with gcc -O2 -Wall fib.c -o fib) are

i=936079, f=610
i=948902, f=8
i=961725, f=233
i=974548, f=3
i=987371, f=89
2.140 cpu sec
./fib  2.15s user 0.00s system 99% cpu 2.152 total

benchmarking a run smaller than about a second is not very meaningful

(and you can use the time command to measure such a run)

^{See also time(7) and clock_gettime(2).}

score 2 · Answer 3 · answered Nov 23 '11 at 09:43

2

It might be easier to use the clock function:

clock_t start = clock();
int r = fib(n);
clock_t end = clock();
printf("Elapsed time: %.2f seconds\n", (double)(end - start) / CLOCKS_PER_SEC);

answered Nov 23 '11 at 09:43

Some programmer dude

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does `clock()` give you time used by processor or time elapsed? – Alessandro Pezzato Nov 23 '11 at 09:46
@AlessandroPezzato According to the manual page (linked to in the answer): The clock() function returns an approximation of processor time used by the program. – Some programmer dude Nov 23 '11 at 09:48
2

`clock` measures CPU time, not wall clock time. That may not be what the question is after. – ibid Nov 23 '11 at 09:50

Basile Starynkevitch · Answer 4 · 2011-11-23T13:25:14.233

1

The resolution of the real time clock is probably not very small (perhaps 10 or 25 milliseconds), and your computation is too short to be significant. You could put your computation inside a loop (e.g. repeating it several thousand times).

You could also consider measuring the CPU time, using the clock function.

You could also use the clock_gettime function to get perhaps better results.

And as other people told you, please ask for all warnings with gcc -Wall and take them into account. If you care after performance (but remember that premature optimization is evil, so get your program right first!) consider enabling optimizations (e.g. gcc -Wall -O2) during compilation.

edited Nov 23 '11 at 13:25

answered Nov 23 '11 at 09:49

Basile Starynkevitch

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To tell the truth,I can put a nice number for Fibonacci function,but even I give it a number,like 50,the result time will still be 0.0000 – Eric_Chen Nov 23 '11 at 13:21
Did you repeat your call to Fibonacci a million time? – Basile Starynkevitch Nov 23 '11 at 13:25

score 0 · Answer 5 · answered Nov 23 '11 at 09:51

This should give you elapsed time:

#include <iostream>
#include <sys/time.h> /* gettimeofday */

int main() {
    /* get begin time */
    timeval begin;
    ::gettimeofday(&begin, 0);
    /* do something... */
    ::usleep(153);
    /* get end time */
    ::timeval current;
    ::gettimeofday(&current, (struct timezone*) 0);
    /* calculate difference */
    double elapsed = (current.tv_sec - begin.tv_sec) + ((current.tv_usec
            - begin.tv_usec) / 1000000.0F);
    /* print it */
    std::cout << elapsed << std::endl;
    return 0;
}

I tried the method in c,and it really worked,but when I used it like the code I posted.It turned out to not work — Eric_Chen, Nov 23 '11 at 13:24

How to use c time in linux to print the function running time?

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