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      1 // This artificial program runs a lot of code.  The exact amount depends on
      2 // the command line -- if an arg "0" is given, it does exactly
      3 // the same amount of work, but using four times as much code.
      4 // If an arg >= 1 is given, the amount of code is multiplied by this arg.
      5 //
      6 // It's a stress test for Valgrind's translation speed;  natively the two
      7 // modes run in about the same time (the I-cache effects aren't big enough
      8 // to make a difference), but under Valgrind the one running more code is
      9 // significantly slower due to the extra translation time.
     10 
     11 // 31 Aug 2015: this only "works" on x86/amd64/s390 by accident; the
     12 // test is essentially kludged.  This "generates" code into memory
     13 // (the mmap'd area) and the executes it.  But historically and even
     14 // after this commit (r15601), the test has been run without
     15 // --smc-check=all or all-non-file.  That just happens to work because
     16 // the "generated" code is never modified, so there's never a
     17 // translated-vs-reality coherence problem.  Really we ought to run
     18 // with the new-as-of-r15601 default --smc-check=all-non-file, but that
     19 // hugely slows it down and makes the results non-comparable with
     20 // pre r15601 results, so instead the .vgperf files now specify the
     21 // old default value --smc-check=stack explicitly.
     22 
     23 
     24 #include <stdio.h>
     25 #include <string.h>
     26 #include <stdlib.h>
     27 #include <assert.h>
     28 #if defined(__mips__)
     29 #include <asm/cachectl.h>
     30 #include <sys/syscall.h>
     31 #endif
     32 #include "tests/sys_mman.h"
     33 
     34 #define FN_SIZE   1280     // Must be big enough to hold the compiled f()
     35                            // and any literal pool that might be used
     36 #define N_LOOPS   20000    // Should be divisible by four
     37 #define RATIO     4        // Ratio of code sizes between the two modes
     38 
     39 int f(int x, int y)
     40 {
     41    int i;
     42    for (i = 0; i < 5000; i++) {
     43       switch (x % 8) {
     44        case 1:  y += 3;
     45        case 2:  y += x;
     46        case 3:  y *= 2;
     47        default: y--;
     48       }
     49    }
     50    return y;
     51 }
     52 
     53 int main(int argc, char* argv[])
     54 {
     55    int h, i, sum1 = 0, sum2 = 0, sum3 = 0, sum4 = 0;
     56    int n_fns, n_reps;
     57 
     58    if (argc <= 1) {
     59       // Mode 1: not so much code
     60       n_fns  = N_LOOPS / RATIO;
     61       n_reps = RATIO;
     62       printf("mode 1: ");
     63    } else {
     64       // Mode 2: lots of code
     65       const int mul = atoi(argv[1]);
     66       if (mul == 0)
     67          n_fns = N_LOOPS;
     68       else
     69          n_fns = N_LOOPS * mul;
     70       n_reps = 1;
     71       printf("mode 1: ");
     72    }
     73    printf("%d copies of f(), %d reps\n", n_fns, n_reps);
     74 
     75    char* a = mmap(0, FN_SIZE * n_fns,
     76                      PROT_EXEC|PROT_WRITE|PROT_READ,
     77                      MAP_PRIVATE|MAP_ANONYMOUS, -1,0);
     78    assert(a != (char*)MAP_FAILED);
     79 
     80    // Make a whole lot of copies of f().  FN_SIZE is much bigger than f()
     81    // will ever be (we hope).
     82    for (i = 0; i < n_fns; i++) {
     83       memcpy(&a[FN_SIZE*i], f, FN_SIZE);
     84    }
     85 
     86 #if defined(__mips__)
     87    syscall(__NR_cacheflush, a, FN_SIZE * n_fns, ICACHE);
     88 #endif
     89 
     90    for (h = 0; h < n_reps; h += 1) {
     91       for (i = 0; i < n_fns; i += 4) {
     92          int(*f1)(int,int) = (void*)&a[FN_SIZE*(i+0)];
     93          int(*f2)(int,int) = (void*)&a[FN_SIZE*(i+1)];
     94          int(*f3)(int,int) = (void*)&a[FN_SIZE*(i+2)];
     95          int(*f4)(int,int) = (void*)&a[FN_SIZE*(i+3)];
     96          sum1 += f1(i+0, n_fns-i+0);
     97          sum2 += f2(i+1, n_fns-i+1);
     98          sum3 += f3(i+2, n_fns-i+2);
     99          sum4 += f4(i+3, n_fns-i+3);
    100          if (i % 1000 == 0)
    101             printf(".");
    102       }
    103    }
    104    printf("result = %d\n", sum1 + sum2 + sum3 + sum4);
    105    return 0;
    106 }
    107