Re: [問題] 使用CUDA來擷取矩陣中部分值

[lgen7604 ()]

我執行的結果正常啊

取上下左右的部份都沒問題

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <cuda_runtime.h>
#include <cutil.h>

#define NNx 4
#define NNy 4

__global__ void movaIM5(float *input1, float *output1)
{
  int col = blockIdx.x*blockDim.x+threadIdx.x;
  int row = blockIdx.y*blockDim.y+threadIdx.y;
  output1[row*NNx+col] = input1[(row+1)*NNx+col]; // 取下面矩陣
  //output1[row*NNx+col] = input1[row*NNx+col]; // 取上面矩陣
}

__global__ void movaIM6(float *input1, float *output1)
{
  int col = blockIdx.x*blockDim.x+threadIdx.x;
  int row = blockIdx.y*blockDim.y+threadIdx.y;
  output1[row*(NNx-1)+col] = input1[row*NNx+col+1];  // 取右邊矩陣
  //output1[row*(NNx-1)+col] = input1[row*NNx+col];  // 取左邊矩陣
}

int main(int argc, char* argv[])
{
  int i;
  float input[NNx*NNy];
  float *d_input;

  for(i = 0; i < NNx*NNy; i++){
    input[i] = i+1;
    if(i % NNx == 0) printf("\n");
    printf("%3.0f ", input[i]);
  }
  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);

  dim3 blocks(1,1);
  dim3 threadsIM5(NNx,NNy-1);
  dim3 threadsIM6(NNx-1,NNy);
  movaIM5<<<blocks, threadsIM5>>>( d_input, d_IM5 );
  movaIM6<<<blocks, threadsIM6>>>( 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 );

  for(i = 0; i < NNx*(NNy-1); i++){
    if(i % NNx == 0) printf("\n");
    printf("%3.0f ", IM5[i]);
  }
  printf("\n");
  for(i = 0; i < (NNx-1)*NNy; i++){
    if(i % (NNx-1) == 0) printf("\n");
    printf("%3.0f ", IM6[i]);
  }
  printf("\n");

  cudaFree(d_input);
  cudaFree(d_IM5);
  cudaFree(d_IM6);
  system("pause");

  return 0;
}

--
※ 發信站: 批踢踢實業坊(ptt.cc) 
◆ From: 122.120.40.234
→ aada:如果我今天矩陣是比較大的畫,如512*512
→ aada:dim3 blocks, threadsIM5好像就不可以這樣寫了,是嗎
→ aada:int bx = (NNx + BLOCK_SIZE - 1) / BLOCK_SIZE;
→ aada:dim3 blocks(bx, bx);
→ aada:dim3 threads(BLOCK_SIZE, BLOCK_SIZE);
→ aada:BLOCK_SIZE 我設為16
推 aada:我IM5大致上試出來了
→ aada:dim3 blocksIM5( 16, 73);
→ aada:dim3 threadsIM5( 32, 7 );
→ aada:movaIM5<<<blocksIM5, threadsIM5>>>( d_input, d_IM5 );
→ aada:這樣剛好處理(16*32=512, 73*7=511) --> 512*511的矩陣
→ aada:我這樣寫的方式OK嗎,有什麼地方可以改進的