我试图更多地了解 CPU 缓存如何影响性能。作为一个简单的测试，我将矩阵第一列的值与不同数量的总列数相加。
// compiled with: gcc -Wall -Wextra -Ofast -march=native cache.c
// tested with: for n in {1..100}; do ./a.out $n; done | tee out.csv
#include <assert.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
double sum_column(uint64_t ni, uint64_t nj, double const data[ni][nj])
{
double sum = 0.0;
for (uint64_t i = 0; i < ni; ++i) {
sum += data[i][0];
}
return sum;
}
int compare(void const* _a, void const* _b)
{
double const a = *((double*)_a);
double …我正在尝试编写一个与numpy.sum双精度数组一样快的 C 程序，但似乎失败了。
以下是我衡量 numpy 性能的方法：
import numpy as np
import time
SIZE=4000000
REPS=5
xs = np.random.rand(SIZE)
print(xs.dtype)
for _ in range(REPS):
start = time.perf_counter()
r = np.sum(xs)
end = time.perf_counter()
print(f"{SIZE / (end-start) / 10**6:.2f} MFLOPS ({r:.2f})")
输出是：
float64
2941.61 MFLOPS (2000279.78)
3083.56 MFLOPS (2000279.78)
3406.18 MFLOPS (2000279.78)
3712.33 MFLOPS (2000279.78)
3661.15 MFLOPS (2000279.78)
现在尝试在 C 中做类似的事情：
float64
2941.61 MFLOPS (2000279.78)
3083.56 MFLOPS (2000279.78)
3406.18 MFLOPS (2000279.78)
3712.33 MFLOPS (2000279.78)
3661.15 MFLOPS (2000279.78)
编译并gcc -o main …