Question

I’m doing a project that involves comparing programming languages. I’m computing the Ackermann function. I tested Java, Python, and Ruby, and got responses between 10 and 30 milliseconds. But C++ seems to take 125 milliseconds. Is this normal, or is it a problem with the gettimeofday()? Gettimeofday() is in time.h.

I’m testing on a (virtual) Ubuntu Natty Narwhal 32-bit. I’m not short processing power (Quad-core 2.13 GHz Intel Xeon).

My code is here:

#include <iostream>
#include <sys/time.h>
using namespace std;
int a(int m,int n) {
    if (m == 0) {
    return n + 1;
    } else if (m > 0 and n == 0) {
    return a(m-1,1);
    } else if (m > 0 and n > 0) {
    return a(m-1,a(m,n-1));
    }
}

int main() {
    timeval tim;
    gettimeofday(&tim,NULL);
    double t1 = tim.tv_usec;
    int v = a(3,4);           
    gettimeofday(&tim,NULL);
    double t2 = tim.tv_usec;
    cout << v << endl << t2-t1;
    return 0;
}       

Answer 1

Assuming you’re talking about the resolution of the data returned, the POSIX specification for gettimeofday states:

The resolution of the system clock is unspecified.

This is due to the fact that systems may have a widely varying capacity for tracking small time periods. Even the ISO standard clock() function includes caveats like this.

If you’re talking about how long it takes to call it, the standard makes no guarantees about performance along those lines. An implementation is perfectly free to wait 125 minutes before giving you the time although I doubt such an implementation would have much market success 🙂

---

As an example of the limited resolution, I typed in the following code to check it on my system:

#include <stdio.h>
#include <sys/time.h>

#define NUMBER 30

int main (void) {
    struct timeval tv[NUMBER];
    int count[NUMBER], i, diff;

    gettimeofday (&tv[0], NULL);

    for (i = 1; i < NUMBER; i++) {
        gettimeofday (&tv[i], NULL);
        count[i] = 1;
        while ((tv[i].tv_sec == tv[i-1].tv_sec) &&
            (tv[i].tv_usec == tv[i-1].tv_usec))
        {
            count[i]++;
            gettimeofday (&tv[i], NULL);
        }
    }

    printf ("%2d: secs = %d, usecs = %6d\n", 0, tv[0].tv_sec, tv[0].tv_usec);
    for (i = 1; i < NUMBER; i++) {
        diff = (tv[i].tv_sec - tv[i-1].tv_sec) * 1000000;
        diff += tv[i].tv_usec - tv[i-1].tv_usec;

        printf ("%2d: secs = %d, usecs = %6d, count = %5d, diff = %d\n",
            i, tv[i].tv_sec, tv[i].tv_usec, count[i], diff);
    }

    return 0;
}

The code basically records the changes in the underlying time, keeping a count of how many calls it took to gettimeofday() for the time to actually change. This is on a reasonably powerful machine so it’s not short on processing power (the count indicates how often it was able to call gettimeofday() for each time quantum, around the 5,800 mark, ignoring the first since we don’t know when in that quantum we started the measurements).

The output was:

 0: secs = 1318554836, usecs = 990820
 1: secs = 1318554836, usecs = 991820, count =  5129, diff = 1000
 2: secs = 1318554836, usecs = 992820, count =  5807, diff = 1000
 3: secs = 1318554836, usecs = 993820, count =  5901, diff = 1000
 4: secs = 1318554836, usecs = 994820, count =  5916, diff = 1000
 5: secs = 1318554836, usecs = 995820, count =  5925, diff = 1000
 6: secs = 1318554836, usecs = 996820, count =  5814, diff = 1000
 7: secs = 1318554836, usecs = 997820, count =  5814, diff = 1000
 8: secs = 1318554836, usecs = 998820, count =  5819, diff = 1000
 9: secs = 1318554836, usecs = 999820, count =  5901, diff = 1000
10: secs = 1318554837, usecs =    820, count =  5815, diff = 1000
11: secs = 1318554837, usecs =   1820, count =  5866, diff = 1000
12: secs = 1318554837, usecs =   2820, count =  5849, diff = 1000
13: secs = 1318554837, usecs =   3820, count =  5857, diff = 1000
14: secs = 1318554837, usecs =   4820, count =  5867, diff = 1000
15: secs = 1318554837, usecs =   5820, count =  5852, diff = 1000
16: secs = 1318554837, usecs =   6820, count =  5865, diff = 1000
17: secs = 1318554837, usecs =   7820, count =  5867, diff = 1000
18: secs = 1318554837, usecs =   8820, count =  5885, diff = 1000
19: secs = 1318554837, usecs =   9820, count =  5864, diff = 1000
20: secs = 1318554837, usecs =  10820, count =  5918, diff = 1000
21: secs = 1318554837, usecs =  11820, count =  5869, diff = 1000
22: secs = 1318554837, usecs =  12820, count =  5866, diff = 1000
23: secs = 1318554837, usecs =  13820, count =  5875, diff = 1000
24: secs = 1318554837, usecs =  14820, count =  5925, diff = 1000
25: secs = 1318554837, usecs =  15820, count =  5870, diff = 1000
26: secs = 1318554837, usecs =  16820, count =  5877, diff = 1000
27: secs = 1318554837, usecs =  17820, count =  5868, diff = 1000
28: secs = 1318554837, usecs =  18820, count =  5874, diff = 1000
29: secs = 1318554837, usecs =  19820, count =  5862, diff = 1000

showing that the resolution seems to be limited to no better than one thousand microseconds. Of course, your system may be different to that, the bottom line is that it depends on your implementation and/or environment.

---

One way to get around this type of limitation is to not do something once but to do it N times and then divide the elapsed time by N.

For example, let’s say you call your function and the timer says it took 125 milliseconds, something that you suspect seems a little high. I would suggest then calling it a thousand times in a loop, measuring the time it took for the entire thousand.

If that turns out to be 125 seconds then, yes, it’s probably slow. However, if it takes only 27 seconds, that would indicate your timer resolution is what’s causing the seemingly large times, since that would equate to 27 milliseconds per iteration, on par with what you’re seeing from the other results.

Modifying your code to take this into account would be along the lines of:

int main() {
    const int count = 1000;
    timeval tim;

    gettimeofday(&tim, NULL);
    double t1 = 1.0e6 * tim.tv_sec + tim.tv_usec;

    int v;
    for (int i = 0; i < count; ++i)
        v = a(3, 4);           

    gettimeofday(&tim, NULL);
    double t2 = 1.0e6 * tim.tv_sec + tim.tv_usec;

    cout << v << '\n' << ((t2 - t1) / count) << '\n';

    return 0;
}