我在最后几天开始和cuda一起工作。编写一个乘以N×N大小的两个矩阵的程序是没有问题的。在内核函数中,我使用了这段代码:
for(int i = 0; i < width; i++){
sum += a[row * width + i] * b[i * width + col];
c[row * width + col] = sum;
}
如何设计核函数以将大小为1 x N的矩阵与大小为N x M的矩阵相乘
答案 0 :(得分:-2)
我现在找到了解决这个问题的方法:
#include <stdio.h>
#include <iostream>
using namespace std;
__global__
void kernel(float *a, float *b, float *c, int N, int M) {
int tid = threadIdx.x + blockIdx.x * blockDim.x;
float sum = 0;
if (tid < M) {
for (int i = 0; i < N; i++)
sum += a[i] * b[(i * M) + tid];
c[tid] = sum;
}
}
int main(void) {
float *dev_a, *dev_b, *dev_c;
int N = 16;
int M = 12;
float a[N];
float b[N][M];
float c[M];
for (int i = 0; i < N; i++) {
a[i] = 1.0;
}
for (int i = 0; i < N; i++) {
for (int e = 0; e < M; e++) {
b[i][e] = 1.0;
}
}
cudaMalloc((void**) &dev_a, sizeof(float) * N);
cudaMalloc((void**) &dev_b, sizeof(float) * N * M);
cudaMalloc((void**) &dev_c, sizeof(float) * M);
cudaMemcpy(dev_a, a, sizeof(float) * N, cudaMemcpyHostToDevice);
cudaMemcpy(dev_b, b, sizeof(float) * N * M, cudaMemcpyHostToDevice);
kernel<<<M / 256 + 1, 256>>>(dev_a, dev_b, dev_c, N, M);
cudaMemcpy(c, dev_c, sizeof(float) * M, cudaMemcpyDeviceToHost);
cudaFree(dev_a);
cudaFree(dev_b);
cudaFree(dev_c);
for (int i = 0; i < M; i++) {
cout << c[i] << endl;
}
return 0;
}
但我还有一个问题。出于性能原因,在几个内核中拆分for循环操作是否有意义?