1 // This artificial program runs a lot of code. The exact amount depends on
2 // the command line -- if any command line args are given, it does exactly
3 // the same amount of work, but using four times as much code.
4 //
5 // It's a stress test for Valgrind's translation speed; natively the two
6 // modes run in about the same time (the I-cache effects aren't big enough
7 // to make a difference), but under Valgrind the one running more code is
8 // significantly slower due to the extra translation time.
9
10 #include <stdio.h>
11 #include <string.h>
12 #include <assert.h>
13 #include "tests/sys_mman.h"
14
15 #define FN_SIZE 996 // Must be big enough to hold the compiled f()
16 #define N_LOOPS 20000 // Should be divisible by four
17 #define RATIO 4 // Ratio of code sizes between the two modes
18
f(int x,int y)19 int f(int x, int y)
20 {
21 int i;
22 for (i = 0; i < 5000; i++) {
23 switch (x % 8) {
24 case 1: y += 3;
25 case 2: y += x;
26 case 3: y *= 2;
27 default: y--;
28 }
29 }
30 return y;
31 }
32
main(int argc,char * argv[])33 int main(int argc, char* argv[])
34 {
35 int h, i, sum1 = 0, sum2 = 0, sum3 = 0, sum4 = 0;
36 int n_fns, n_reps;
37
38 char* a = mmap(0, FN_SIZE * N_LOOPS,
39 PROT_EXEC|PROT_WRITE,
40 MAP_PRIVATE|MAP_ANONYMOUS, -1,0);
41 assert(a != (char*)MAP_FAILED);
42
43 if (argc <= 1) {
44 // Mode 1: not so much code
45 n_fns = N_LOOPS / RATIO;
46 n_reps = RATIO;
47 printf("mode 1: ");
48 } else {
49 // Mode 2: lots of code
50 n_fns = N_LOOPS;
51 n_reps = 1;
52 printf("mode 1: ");
53 }
54 printf("%d copies of f(), %d reps\n", n_fns, n_reps);
55
56 // Make a whole lot of copies of f(). FN_SIZE is much bigger than f()
57 // will ever be (we hope).
58 for (i = 0; i < n_fns; i++) {
59 memcpy(&a[FN_SIZE*i], f, FN_SIZE);
60 }
61
62 for (h = 0; h < n_reps; h += 1) {
63 for (i = 0; i < n_fns; i += 4) {
64 int(*f1)(int,int) = (void*)&a[FN_SIZE*(i+0)];
65 int(*f2)(int,int) = (void*)&a[FN_SIZE*(i+1)];
66 int(*f3)(int,int) = (void*)&a[FN_SIZE*(i+2)];
67 int(*f4)(int,int) = (void*)&a[FN_SIZE*(i+3)];
68 sum1 += f1(i+0, n_fns-i+0);
69 sum2 += f2(i+1, n_fns-i+1);
70 sum3 += f3(i+2, n_fns-i+2);
71 sum4 += f4(i+3, n_fns-i+3);
72 if (i % 1000 == 0)
73 printf(".");
74 }
75 }
76 printf("result = %d\n", sum1 + sum2 + sum3 + sum4);
77 return 0;
78 }
79