c - 分析C程序函数的缓存命中率

Question

我想获得在 Linux 机器上运行的 C/C++ 程序 (foo) 的特定函数的缓存命中率。我正在使用 gcc 并且没有编译器优化。借助 perf，我可以使用以下命令获得整个程序的命中率。

perf stat -e L1-dcache-loads,L1-dcache-load-misses,L1-dcache-stores,L1-dcache-store-misses ./a.out

但我只对内核 foo 感兴趣。

有没有一种方法可以使用 perf 或任何其他工具只获得 foo 的命中率？

#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>


#define NI 192
#define NJ NI

#ifndef DATA_TYPE
    #define DATA_TYPE float
#endif


static 
void* xmalloc(size_t num)
{
    void * nnew = NULL;
    int ret = posix_memalign (&nnew, 32, num);
    if(!nnew || ret)
    {
        fprintf(stderr, "Can not allocate Memory\n");
        exit(1);
    }
    return nnew;
}

void* alloc_data(unsigned long long int n, int elt_size)
{
    size_t val = n;
    val *= elt_size;
    void* ret = xmalloc(val);
    return ret;
}


/* Array initialization. */
static
void init_array(int ni, int nj,
        DATA_TYPE A[NI][NJ],
        DATA_TYPE R[NJ][NJ],
        DATA_TYPE Q[NI][NJ])
{
  int i, j;

  for (i = 0; i < ni; i++)
    for (j = 0; j < nj; j++) {
      A[i][j] = ((DATA_TYPE) i*j) / ni;
      Q[i][j] = ((DATA_TYPE) i*(j+1)) / nj;
    }
  for (i = 0; i < nj; i++)
    for (j = 0; j < nj; j++)
      R[i][j] = ((DATA_TYPE) i*(j+2)) / nj;
}


/* Main computational kernel.*/

static
void foo(int ni, int nj,
        DATA_TYPE A[NI][NJ],
        DATA_TYPE R[NJ][NJ],
        DATA_TYPE Q[NI][NJ])
{
  int i, j, k;

  DATA_TYPE nrm;
  for (k = 0; k < nj; k++)
  {
    nrm = 0;
    for (i = 0; i < ni; i++)
      nrm += A[i][k] * A[i][k];
    R[k][k] = sqrt(nrm);
    for (i = 0; i < ni; i++)
      Q[i][k] = A[i][k] / R[k][k];
    for (j = k + 1; j < nj; j++)
    {
      R[k][j] = 0;
      for (i = 0; i < ni; i++)
        R[k][j] += Q[i][k] * A[i][j];
      for (i = 0; i < ni; i++)
        A[i][j] = A[i][j] - Q[i][k] * R[k][j];
    }
  }
}


int main(int argc, char** argv)
{
  /* Retrieve problem size. */
  int ni = NI;
  int nj = NJ;

  /* Variable declaration/allocation. */
  DATA_TYPE (*A)[NI][NJ];
  DATA_TYPE (*R)[NI][NJ];
  DATA_TYPE (*Q)[NI][NJ];

  A = ((DATA_TYPE (*)[NI][NJ])(alloc_data((NI*NJ), (sizeof(DATA_TYPE)))));
  R = ((DATA_TYPE (*)[NI][NJ])(alloc_data((NI*NJ), (sizeof(DATA_TYPE)))));
  Q = ((DATA_TYPE (*)[NI][NJ])(alloc_data((NI*NJ), (sizeof(DATA_TYPE)))));
  
/* Initialize array(s). */
  init_array (ni, nj,
          (*A),
          (*R),
          (*Q));


  /* Run kernel. */
  foo (ni, nj, *A, *R, *Q);

  /* Be clean. */
  free((void *)A);
  free((void *)R);
  free((void *)Q);

  return 0;
}

lscpu 命令的输出是:

Architecture:          x86_64
CPU op-mode(s):        32-bit, 64-bit
Byte Order:            Little Endian
CPU(s):                16
On-line CPU(s) list:   0-15 
Thread(s) per core:    2
Core(s) per socket:    8
Socket(s):             1
NUMA node(s):          1
Vendor ID:             GenuineIntel 
CPU family:            6
Model:                 63
Model name:            Intel(R) Core(TM) i7-5960X CPU @ 3.00GHz
Stepping:              2
CPU max MHz:           3500.0000
CPU min MHz:           1200.0000
L1d cache:             32K
L1i cache:             32K
L2 cache:              256K
L3 cache:              20480K
NUMA node0 CPU(s):     0-15

Answer 1

你也可以使用 Likwid及其 Marker-API .它使检测代码的某些区域变得非常容易。您可以在 haswell 架构上使用预定义的性能组 ICACHE 来获得 L1 缓存未命中率，或者为 L1 命中率定义您自己的性能组。

#include likwid.h
LIKWID_MARKER_INIT;
LIKWID_MARKER_START("region foo");

foo();

LIKWID_MARKER_STOP("region foo");
LIKWID_MARKER_CLOSE;

运行应用程序:

./likwid-perfctr -g ICACHE -m <your application>

确保使用 -DLIKWID-PERFMON 进行编译并添加 Likwid 包含和库路径并链接 Likwid 库:-L$LIKWID_LIB -I$LIKWID_INCLUDE -llikwid .一切都在他们的 github wiki 上得到了很好的记录。