callbench.c

// gcc callbench.c -o callbench -O1
#include <stdio.h>
#include <stdlib.h>
#include "cycle.h"

int f0( int x )
{
    /* The details of this calculation are not important */
    return x;
}

int f1( int x) {
	return f0 ( f0 ( x));
}

int f2( int x) {
	return f1 ( f1 ( x));
}

int f3( int x) {
	return f2 ( f2 ( x));
}

int f4( int x) {
	return f3 ( f3 ( x));
}

int f5( int x) {
	return f4 ( f4 ( x));
}

int f6( int x) {
	return f5 ( f5 ( x));
}

int f7( int x) {
	return f6 ( f6 ( x));
}

int f8( int x) {
	return f7 ( f7 ( x));
}

int f9( int x) {
	return f8 ( f8 ( x));
}

int f10( int x) {
	return f9 ( f9 ( x));
}

int f11( int x) {
	return f10 ( f10 ( x));
}

int f12( int x) {
	return f11 ( f11 ( x));
}

int f13( int x) {
	return f12 ( f12 ( x));
}

int f14( int x) {
	return f13 ( f13 ( x));
}

int f15( int x) {
	return f14 ( f14 ( x));
}

int f16( int x) {
	return f15 ( f15( x));
}

int main (int argc, char *argv[]) {

	int iterations = atoi(argv[1]);

	ticks start = getticks();

	int result = 0;
	int i = iterations;
	while (i --> 0) {
		result += f16 (i);
	}

	ticks end = getticks();

	double total = elapsed(end, start);
	/**
	 * f16 will be called once, f15 twice, f14 four times, ... f0 2^16 times
	 * so the total number of function calls is 2^17 - 1.
	 */
	double percall = total / ( (1 << 17) - 1) / iterations;
	printf("total cycles: %.2f, cycles per call: %.2f\n", total, percall);

	return result;
}