CUDA共享内存的可能性

Question

I'm having a bit of a issue understanding how I could implement CUDA shared memory since I'm not using the thread number for anything other than to check which calculations that certain thread should do.

__global__ void gpu_histogram_equalization(unsigned char * img_out, unsigned char * img_in,
                            int * hist_in, int img_size, int nbr_bin, int numOfThreads, int * lut){


    int i = 0;
    int x = threadIdx.x + blockDim.x*blockIdx.x;

    int start;
    int end;

    /* Get the result image */
    if(x >= img_size) {
       return;
    }
    start = ((img_size/numOfThreads) * x);
    if(numOfThreads == 1) {
       end = (img_size/numOfThreads);
    }
    else {
       end = ((img_size/numOfThreads) * (x+1));
    }
    for(i = start; i < end; i ++){
        if(lut[img_in[i]] > 255){
            img_out[i] = 255;
        }
        else{
            img_out[i] = (unsigned char)lut[img_in[i]];
        }

    }
}

Can anyone clarify that my speculation is true, that this is not possible to make use of shared memory?

Answer 1

Using shared memory will give you a preformance increase if you reuse the data multiple times. The code can be rewritten to utilize higher memory bandwidth and discard the use of shared memory.

Something like this:

__global__ void gpu_histogram_equalization(unsigned char * img_out, unsigned char * img_in,
                        int * hist_in, int img_size, int nbr_bin, int numOfThreads, int * lut){
  int lutval;
  int x = threadIdx.x + blockDim.x*blockIdx.x;

  /* Get the result image */
  if(x >= img_size) {
     return;
  }

  lutval = lut[img_in[x]];

  if(lutval > 255){
   img_out[x] = 255;
  }
  else{
    img_out[i] = (unsigned char)lutval;
  }
}