Homework6 向量乘法

22 May 2019 1393字 5分
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题目链接

给出以下 CUDA 程序（点击参考代码获取），补充完整代码。 需手动切换编程语言为 CUDA。 输出：

0.000000 0.000000 0.000000
0.100000 -0.100000 -0.010000
0.200000 -0.200000 -0.040000
0.300000 -0.300000 -0.090000
0.400000 -0.400000 -0.160000
0.500000 -0.500000 -0.250000

心中毫无波澜…

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

__global__ void multiply(float *A, float *B, float *C, const int N)
{
	//Your code here.
	for (int i = threadIdx.x + blockIdx.x * blockDim.x; i < N; i += blockDim.x * gridDim.x)
		C[i] = A[i] * B[i];
	//End of your code.
}

int main()
{
	int nElem = 6;
	size_t nBytes = nElem * sizeof(float);
	float *h_A, *h_B, *h_C;

	h_A = (float *)malloc(nBytes);
	h_B = (float *)malloc(nBytes);
	h_C = (float *)malloc(nBytes);

	memset(h_A, 0, nBytes);
	memset(h_B, 0, nBytes);
	memset(h_C, 0, nBytes);

	for (int i = 0; i < nElem; i++)
	{
		h_A[i] = i * 0.1;
		h_B[i] = -i * 0.1;
	}

	float *d_A, *d_B, *d_C;
	cudaMalloc((float **)&d_A, nBytes);
	cudaMalloc((float **)&d_B, nBytes);
	cudaMalloc((float **)&d_C, nBytes);

	cudaMemset(d_A, 0, nBytes);
	cudaMemset(d_B, 0, nBytes);
	cudaMemset(d_C, 0, nBytes);

	cudaMemcpy(d_A, h_A, nBytes, cudaMemcpyHostToDevice);
	cudaMemcpy(d_B, h_B, nBytes, cudaMemcpyHostToDevice);

	multiply<<<2, 3>>>(d_A, d_B, d_C, nElem);

	// Your code here.
	cudaMemcpy(h_C, d_C, nBytes, cudaMemcpyDeviceToHost);
	// End of your code.

	for (int i = 0; i < nElem; i++)
		printf("%f\t%f\t%f\n", h_A[i], h_B[i], h_C[i]);

	free(h_A);
	free(h_B);
	free(h_C);

	cudaFree(d_A);
	cudaFree(d_B);
	cudaFree(d_C);
	return 0;
}