作者lgen7604 ()
看板C_and_CPP
标题Re: [问题] 使用CUDA来撷取矩阵中部分值
时间Fri Feb 26 10:19:41 2010
※ 引述《aada (aada)》之铭言：
从你的描述来看你是想把矩阵A
(NNy x NNx) 撷取出一部份
(NNy x NNx-1)
再放入新的矩阵B
(NNy x NNx+1)
请特别注意矩阵的维度 还有矩阵的row,col及CUDA中x,y的关系
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <cuda_runtime.h>
#include <cutil.h>
#define NNx 16
#define NNy 16
#define BLOCK_SIZE 16
#define ALIGN 3 // 对齐printf用
__global__ void movaIM5(float *input1, float *output1)
{
int col = blockIdx.x*blockDim.x+threadIdx.x;
int row = blockIdx.y*blockDim.y+threadIdx.y;
if(col+1 < NNx && row < NNy)
output1[row*(NNx-1)+col] = input1[row*NNx+col+1]; // 取右边矩阵
//if(col < (NNx-1) && row < NNy)
// output1[row*(NNx-1)+col] = input1[row*NNx+col]; // 取左边矩阵
}
__global__ void movaIM6(float *input1, float *output1)
{
int col = blockIdx.x*blockDim.x+threadIdx.x;
int row = blockIdx.y*blockDim.y+threadIdx.y;
if(col < NNx && row+1 < NNy)
output1[row*NNx+col] = input1[(row+1)*NNx+col]; // 取下面矩阵
//if(col < NNx && row < (NNy-1))
// output1[row*NNx+col] = input1[row*NNx+col]; // 取上面矩阵
}
__global__ void cu_delx( float *input1, float *input2 )
{
int col = blockIdx.x*blockDim.x+threadIdx.x;
int row = blockIdx.y*blockDim.y+threadIdx.y;
input1[row*NNx+col]=0;
// 因为你要搬的矩阵是 (NNy x NNx-1)
if( col < NNx-1 && row < NNy )
{
//input1[ row*NNx+col +1+threadIdx.y] =input2[ row*NNx+col ];
input1[ row*(NNx+1)+col+1] =input2[ row*(NNx-1)+col ];
}
}
int main(int argc, char* argv[])
{
int i;
//float input[NNx*NNy];
float *input;
input = (float*) malloc( sizeof(float)*NNx*NNy );
float *d_input;
printf("input\n");
for(i = 0; i < NNx*NNy; i++)
{
input[i] = i+1;
}
printf("\n");
cudaMalloc((void**)&d_input, sizeof(float)*NNx*NNy);
cudaMemcpy( d_input, input, sizeof(float)*NNx*NNy, cudaMemcpyHostToDevice );
float *d_IM5, *d_IM6;
cudaMalloc((void**)&d_IM5, sizeof(float)*NNx*(NNy-1));
cudaMalloc((void**)&d_IM6, sizeof(float)*(NNx-1)*NNy);
int bx = (NNx + BLOCK_SIZE - 1) / BLOCK_SIZE;
int by = (NNy + BLOCK_SIZE - 1) / BLOCK_SIZE;
dim3 blocks(bx,by);
dim3 threads(BLOCK_SIZE,BLOCK_SIZE);
movaIM5<<<blocks, threads>>>( d_input, d_IM5 );
movaIM6<<<blocks, threads>>>( d_input, d_IM6 );
float *IM5, *IM6;
IM5 = (float*) malloc( sizeof(float)*NNx*(NNy-1) );
IM6 = (float*) malloc( sizeof(float)*(NNx-1)*NNy );
cudaMemcpy( IM5, d_IM5, sizeof(float)*NNx*(NNy-1), cudaMemcpyDeviceToHost );
cudaMemcpy( IM6, d_IM6, sizeof(float)*(NNx-1)*NNy, cudaMemcpyDeviceToHost );
// block和thread应该设多少 要看你想处理的data
// 为什麽要多设thread? 不懂
int bx_del = ((NNx-1) + BLOCK_SIZE - 1) / BLOCK_SIZE;
int by_del = (NNy + BLOCK_SIZE - 1) / BLOCK_SIZE;
dim3 blocks_del(bx_del,by_del);
dim3 threads_del(BLOCK_SIZE, BLOCK_SIZE);
float *d_delx;
cudaMalloc((void**)&d_delx, sizeof(float)*(NNx+1)*NNy);
cu_delx<<<blocks_del, threads_del>>>( d_delx, d_IM5 );
float *delx;
delx = (float*) malloc( sizeof(float)*(NNx+1)*NNy );
cudaMemcpy( delx, d_delx, sizeof(float)*(NNx+1)*NNy, cudaMemcpyDeviceToHost);
printf("delx\n");
for(i = 0; i < (NNx+1)*NNy; i++)
{
if(i %
(NNx+1) == 0) printf("\n");
else printf(" ");
printf("%*.0f", ALIGN, delx[i]);
}
printf("\n");
cudaFree(d_input);
cudaFree(d_IM5);
cudaFree(d_IM6);
cudaFree(d_delx);
system("pause");
return 0;
}
--
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1F：→ aada:嗯, 我大致上了解了, 谢谢你 02/28 00:15