阵列迭代中的CPU空间缓存位置

我对一级缓存的理解是,内存提取会加载缓存线。假设缓存线大小为64字节,如果我访问地址处的内存 p ,它将从加载整个块 p 到 p + 64 进入缓存。因此,最好从左到右(而不是从右到左)遍历数组,以最大化缓存位置。

然而,我编写了示例C代码,该代码分配了一个一亿字符的数组,将随机值写入其中并求和(复制如下以供参考)。一个版本的代码从左到右求和,另一个版本的代码从右到左求和。当我对其进行基准测试时,我得到了非常相似的结果(其中“时钟周期”是根据 clock . 代码编译时没有进行任何优化。

所以我的问题是:现代处理器除了“缓存读+64字节”之外,还能做些什么吗?它们会前后缓存吗?编译器能否“告诉”处理器代码正在向后迭代?

作为参考,我正在运行 Mac OS X 10.13.3 使用 gcc-7 (Homebrew GCC 7.2.0_1) 7.2.0 以及具有64字节缓存线的x86-64 Intel处理器。

基准:

$ ./a.out
Backward Iterating...took 150101 clock cycles

$ ./a.out
Forward Iterating...took 146545 clock cycles

我预计前向迭代的速度会快64倍,因为每64个元素都应该是缓存命中,而对于后向迭代,每个元素都应该是缓存未命中。

所以,我打电话给cachegrind。两者的缓存命中率几乎相同:

# Left to right iteration
==21773==
==21773== I   refs:      4,006,996,067
==21773== I1  misses:            5,183
==21773== LLi misses:            3,019
==21773== I1  miss rate:          0.00%
==21773== LLi miss rate:          0.00%
==21773==
==21773== D   refs:      1,802,393,260  (1,401,627,925 rd   + 400,765,335 wr)
==21773== D1  misses:        3,153,101  (    1,588,104 rd   +   1,564,997 wr)
==21773== LLd misses:        3,004,885  (    1,440,660 rd   +   1,564,225 wr)
==21773== D1  miss rate:           0.2% (          0.1%     +         0.4%  )
==21773== LLd miss rate:           0.2% (          0.1%     +         0.4%  )
==21773==
==21773== LL refs:           3,158,284  (    1,593,287 rd   +   1,564,997 wr)
==21773== LL misses:         3,007,904  (    1,443,679 rd   +   1,564,225 wr)
==21773== LL miss rate:            0.1% (          0.0%     +         0.4%  )

# Right to left iteration
==21931==
==21931== I   refs:      4,006,996,453
==21931== I1  misses:            5,198
==21931== LLi misses:            3,045
==21931== I1  miss rate:          0.00%
==21931== LLi miss rate:          0.00%
==21931==
==21931== D   refs:      1,802,393,428  (1,401,628,038 rd   + 400,765,390 wr)
==21931== D1  misses:        3,153,113  (    1,588,079 rd   +   1,565,034 wr)
==21931== LLd misses:        3,135,505  (    1,571,219 rd   +   1,564,286 wr)
==21931== D1  miss rate:           0.2% (          0.1%     +         0.4%  )
==21931== LLd miss rate:           0.2% (          0.1%     +         0.4%  )
==21931==
==21931== LL refs:           3,158,311  (    1,593,277 rd   +   1,565,034 wr)
==21931== LL misses:         3,138,550  (    1,574,264 rd   +   1,564,286 wr)
==21931== LL miss rate:            0.1% (          0.0%     +         0.4%  )

代码:

#include <stdint.h>
#include <time.h>
#include <stdio.h>
#include <stdlib.h>

#define BUF_SIZE 100000000

int main() {
  srand(time(NULL));
  uint8_t *buf1 = (uint8_t *)malloc(BUF_SIZE);
  // Fill the buf with random data
  for (size_t i = 0; i < BUF_SIZE; ++i) {
    buf1[i] = rand();
  }

#ifdef BACKWARDS
  printf("Backward Iterating...");
#else
  printf("Forward Iterating...");
#endif

  uint64_t sum = 0;
  clock_t start = clock();
#ifdef BACKWARDS
  for (size_t i = BUF_SIZE - 1; i != ~0; --i) {
#else
  for (size_t i = 0; i < BUF_SIZE; ++i) {
#endif
    sum += buf1[i];
  }
  clock_t end = clock();
  printf("took %lu clock cycles\n", end - start);
  printf("sum: %llu\n", sum);
  free(buf1);
}