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 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 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
