我正在尝试用C程序来理解CPU缓存和缓存行,这是我学习大多数C概念的方式。我使用的程序如下,灵感来自博客:
http://igoro.com/archive/gallery-of-processor-cache-effects/
现在,在我的机器上,使用CFLAGS="-g -O0 -Wall"编译得到的输出如下:
优化后的输出结果 (
在博客中指出,第一个循环将数组中的每个值乘以3,第二个循环仅将每16个值乘以3。第二个循环仅执行了大约6%的第一个循环工作量,但在现代计算机上,这两个for循环所需时间相同:在我的计算机上分别为80ms和78ms。
在我的计算机上似乎不是这种情况。正如您所看到的,执行第一个循环所需的时间是第二个循环的两倍以上。
./cache
CPU time for loop 1 0.460000 secs.
CPU time for loop 2 (j = 8) 0.050000 secs.
CPU time for loop 2 (j = 9) 0.050000 secs.
CPU time for loop 2 (j = 10) 0.050000 secs.
CPU time for loop 2 (j = 11) 0.050000 secs.
CPU time for loop 2 (j = 12) 0.040000 secs.
CPU time for loop 2 (j = 13) 0.050000 secs.
CPU time for loop 2 (j = 14) 0.050000 secs.
CPU time for loop 2 (j = 15) 0.040000 secs.
CPU time for loop 2 (j = 16) 0.050000 secs.
CPU time for loop 2 (j = 17) 0.040000 secs.
CPU time for loop 2 (j = 18) 0.050000 secs.
CPU time for loop 2 (j = 19) 0.040000 secs.
CPU time for loop 2 (j = 20) 0.040000 secs.
CPU time for loop 2 (j = 21) 0.040000 secs.
CPU time for loop 2 (j = 22) 0.040000 secs.
CPU time for loop 2 (j = 23) 0.040000 secs.
CPU time for loop 2 (j = 24) 0.030000 secs.
CPU time for loop 2 (j = 25) 0.040000 secs.
CPU time for loop 2 (j = 26) 0.030000 secs.
CPU time for loop 2 (j = 27) 0.040000 secs.
CPU time for loop 2 (j = 28) 0.030000 secs.
CPU time for loop 2 (j = 29) 0.040000 secs.
CPU time for loop 2 (j = 30) 0.030000 secs.
CPU time for loop 2 (j = 31) 0.030000 secs.
优化后的输出结果 (
CFLAGS=-g -O3 -Wall
)CPU time for loop 1 0.130000 secs.
CPU time for loop 2 (j = 8) 0.040000 secs.
CPU time for loop 2 (j = 9) 0.050000 secs.
CPU time for loop 2 (j = 10) 0.050000 secs.
CPU time for loop 2 (j = 11) 0.040000 secs.
CPU time for loop 2 (j = 12) 0.040000 secs.
CPU time for loop 2 (j = 13) 0.050000 secs.
CPU time for loop 2 (j = 14) 0.050000 secs.
CPU time for loop 2 (j = 15) 0.040000 secs.
CPU time for loop 2 (j = 16) 0.040000 secs.
CPU time for loop 2 (j = 17) 0.050000 secs.
CPU time for loop 2 (j = 18) 0.040000 secs.
CPU time for loop 2 (j = 19) 0.050000 secs.
CPU time for loop 2 (j = 20) 0.040000 secs.
CPU time for loop 2 (j = 21) 0.040000 secs.
CPU time for loop 2 (j = 22) 0.040000 secs.
CPU time for loop 2 (j = 23) 0.030000 secs.
CPU time for loop 2 (j = 24) 0.040000 secs.
CPU time for loop 2 (j = 25) 0.030000 secs.
CPU time for loop 2 (j = 26) 0.040000 secs.
CPU time for loop 2 (j = 27) 0.030000 secs.
CPU time for loop 2 (j = 28) 0.030000 secs.
CPU time for loop 2 (j = 29) 0.030000 secs.
CPU time for loop 2 (j = 30) 0.030000 secs.
CPU time for loop 2 (j = 31) 0.030000 secs.
在博客中指出,第一个循环将数组中的每个值乘以3,第二个循环仅将每16个值乘以3。第二个循环仅执行了大约6%的第一个循环工作量,但在现代计算机上,这两个for循环所需时间相同:在我的计算机上分别为80ms和78ms。
在我的计算机上似乎不是这种情况。正如您所看到的,执行第一个循环所需的时间是第二个循环的两倍以上。
loop 1 is 0.46 seconds.
并且那是针对
loop 2 is 0.03 seconds or 0.04 seconds or 0.05 seconds
针对不同的j值,为什么会出现这种情况?
为什么会发生这种情况?
#include <stdio.h>
#include <sys/time.h>
#include <time.h>
#include <unistd.h>
#include <stdlib.h>
#define MAX_SIZE (64*1024*1024)
int main()
{
clock_t start, end;
double cpu_time;
int i = 0;
int j = 0;
/* MAX_SIZE array is too big for stack. This is an unfortunate rough edge of the way the stack works.
It lives in a fixed-size buffer, set by the program executable's configuration according to the
operating system, but its actual size is seldom checked against the available space. */
/* int arr[MAX_SIZE]; */
int *arr = (int*)malloc(MAX_SIZE * sizeof(int));
/* CPU clock ticks count start */
start = clock();
/* Loop 1 */
for (i = 0; i < MAX_SIZE; i++)
arr[i] *= 3;
/* CPU clock ticks count stop */
end = clock();
cpu_time = ((double) (end - start)) / CLOCKS_PER_SEC;
printf("CPU time for loop 1 %.6f secs.\n", cpu_time);
for (j = 8 ; j < 32 ; j++)
{
/* CPU clock ticks count start */
start = clock();
/* Loop 2 */
for (i = 0; i < MAX_SIZE; i += j)
arr[i] *= 3;
/* CPU clock ticks count stop*/
end = clock();
cpu_time = ((double) (end - start)) / CLOCKS_PER_SEC;
printf("CPU time for loop 2 (j = %d) %.6f secs.\n", j, cpu_time);
}
return 0;
}