blob: 6e8940140bfed38b4de75e624a9f06104b2d3fbe [file] [log] [blame]
// This artificial program runs a lot of code. The exact amount depends on
// the command line -- if an arg "0" is given, it does exactly
// the same amount of work, but using four times as much code.
// If an arg >= 1 is given, the amount of code is multiplied by this arg.
//
// It's a stress test for Valgrind's translation speed; natively the two
// modes run in about the same time (the I-cache effects aren't big enough
// to make a difference), but under Valgrind the one running more code is
// significantly slower due to the extra translation time.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#if defined(__mips__)
#include <asm/cachectl.h>
#include <sys/syscall.h>
#endif
#include "tests/sys_mman.h"
#define FN_SIZE 996 // Must be big enough to hold the compiled f()
#define N_LOOPS 20000 // Should be divisible by four
#define RATIO 4 // Ratio of code sizes between the two modes
int f(int x, int y)
{
int i;
for (i = 0; i < 5000; i++) {
switch (x % 8) {
case 1: y += 3;
case 2: y += x;
case 3: y *= 2;
default: y--;
}
}
return y;
}
int main(int argc, char* argv[])
{
int h, i, sum1 = 0, sum2 = 0, sum3 = 0, sum4 = 0;
int n_fns, n_reps;
if (argc <= 1) {
// Mode 1: not so much code
n_fns = N_LOOPS / RATIO;
n_reps = RATIO;
printf("mode 1: ");
} else {
// Mode 2: lots of code
const int mul = atoi(argv[1]);
if (mul == 0)
n_fns = N_LOOPS;
else
n_fns = N_LOOPS * mul;
n_reps = 1;
printf("mode 1: ");
}
printf("%d copies of f(), %d reps\n", n_fns, n_reps);
char* a = mmap(0, FN_SIZE * n_fns,
PROT_EXEC|PROT_WRITE,
MAP_PRIVATE|MAP_ANONYMOUS, -1,0);
assert(a != (char*)MAP_FAILED);
// Make a whole lot of copies of f(). FN_SIZE is much bigger than f()
// will ever be (we hope).
for (i = 0; i < n_fns; i++) {
memcpy(&a[FN_SIZE*i], f, FN_SIZE);
}
#if defined(__mips__)
syscall(__NR_cacheflush, a, FN_SIZE * n_fns, ICACHE);
#endif
for (h = 0; h < n_reps; h += 1) {
for (i = 0; i < n_fns; i += 4) {
int(*f1)(int,int) = (void*)&a[FN_SIZE*(i+0)];
int(*f2)(int,int) = (void*)&a[FN_SIZE*(i+1)];
int(*f3)(int,int) = (void*)&a[FN_SIZE*(i+2)];
int(*f4)(int,int) = (void*)&a[FN_SIZE*(i+3)];
sum1 += f1(i+0, n_fns-i+0);
sum2 += f2(i+1, n_fns-i+1);
sum3 += f3(i+2, n_fns-i+2);
sum4 += f4(i+3, n_fns-i+3);
if (i % 1000 == 0)
printf(".");
}
}
printf("result = %d\n", sum1 + sum2 + sum3 + sum4);
return 0;
}