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paz4r2p6k 9 months ago
parent 1caeeb357d
commit 970f7bc84e

@ -0,0 +1,104 @@
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <arm_neon.h>
void matmul(float** A, float** B, float** C, int n) {
int i,k,j;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
C[i][j] = 0;
for (k = 0; k < n; k++) {
C[i][j] += A[i][k] * B[k][j];
}
}
}
}
// NEON优化的矩阵乘法函数
void matmul_optimized(float** A, float** B, float** C, int n) {
int i,j,k;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
// 初始化结果矩阵C的当前元素为0
C[i][j] = 0;
// 用于累加的向量寄存器初始化为0
float32x4_t vecC = vdupq_n_f32(0);
for (k = 0; k < n; k += 4) {
// 加载矩阵A的一行中的4个连续元素到向量寄存器
float32x4_t vecA = vld1q_f32(&A[i][k]);
// 加载矩阵B的一列中的4个连续元素到向量寄存器注意这里要转置的逻辑实际是按列取元素
float32x4_t vecB = vld1q_f32(&B[k][j]);
// 对应元素相乘并累加到vecC
vecC = vmlaq_f32(vecC, vecA, vecB);
}
// 将累加结果从向量寄存器提取并累加到C[i][j]
C[i][j] += vgetq_lane_f32(vecC, 0) + vgetq_lane_f32(vecC, 1) + vgetq_lane_f32(vecC, 2) + vgetq_lane_f32(vecC, 3);
}
}
}
int main() {
// 定义矩阵大小
int n = 1024;
// 动态分配两个输入矩阵A和B以及结果矩阵C的内存
float** A = (float**)malloc(n * sizeof(float*));
float** B = (float**)malloc(n * sizeof(float*));
float** C = (float**)malloc(n * sizeof(float*));
int i,j;
for (i = 0; i < n; i++) {
A[i] = (float*)malloc(n * sizeof(float));
B[i] = (float*)malloc(n * sizeof(float));
C[i] = (float*)malloc(n * sizeof(float));
}
// 初始化矩阵数据将A和B矩阵的每个元素随机初始化
srand((unsigned int)time(NULL));
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
A[i][j] = (float)(rand() % 100);
B[i][j] = (float)(rand() % 100);
}
}
clock_t start, end;
// 测试普通矩阵乘法函数的运行时间
start = clock();
matmul(A, B, C, n);
end = clock();
double time_taken_normal = ((double)(end - start)) / CLOCKS_PER_SEC;
printf("normal time: %lf s\n", time_taken_normal);
// 重新初始化结果矩阵C为0以便再次使用
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
C[i][j] = 0;
}
}
// 测试NEON优化矩阵乘法函数的运行时间
start = clock();
matmul_optimized(A, B, C, n);
end = clock();
double time_taken_optimized = ((double)(end - start)) / CLOCKS_PER_SEC;
printf("NEON time: %lf s\n", time_taken_optimized);
// 释放动态分配的内存空间
for ( i = 0; i < n; i++) {
free(A[i]);
free(B[i]);
free(C[i]);
}
free(A);
free(B);
free(C);
return 0;
}
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