How to occupy 80% CPU consistently?

Question

I'm looking for a way to occupy exactly 80% (or any other number) of a single CPU in a consistent manner.
I need this for some unit test that tests a component that triggers under specific CPU utilization conditions
For this purpose I can assume that the machine is otherwise idle.

What's a robust and possibly OS independent way to do this?

Answer 1

There is no such thing as occupying the CPU 80% of the time. The CPU is always either being used, or idle. Over some period of time, you can get average usage to be 80%. Is there a specific time period you want it to be averaged over? This pseudo-code should work across platforms and over 1 second have a CPU usage of 80%:

while True:
    startTime = time.now()
    while date.now() - startTime < 0.8:
        Math.factorial(100) // Or any other computation here
    time.sleep(0.2)

Answer 2

Its pretty easy to write a program that alternately spins and sleeps to get any particular load level you want. I threw this together in a couple of minutes:

#include <stdlib.h>
#include <signal.h>
#include <string.h>
#include <time.h>
#include <sys/time.h>

#define INTERVAL    500000
volatile sig_atomic_t   flag;
void setflag(int sig) { flag = 1; }

int main(int ac, char **av) {
    int load = 80;
    struct sigaction sigact;
    struct itimerval interval = { { 0, INTERVAL }, { 0, INTERVAL } };
    struct timespec pausetime = { 0, 0 };
    memset(&sigact, 0, sizeof(sigact));
    sigact.sa_handler = setflag;
    sigaction(SIGALRM, &sigact, 0);
    setitimer(ITIMER_REAL, &interval, 0);
    if (ac == 2) load = atoi(av[1]);
    pausetime.tv_nsec = INTERVAL*(100 - load)*10;
    while (1) {
        flag = 0;
        nanosleep(&pausetime, 0);
        while (!flag) { /* spin */ } }
    return 0;
}