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CUDA kernel - nested for loop

Tags:

cuda

Hello I'm trying to write a CUDA kernel to perform the following piece of code.

for (n = 0; n < (total-1); n++)
{
  a = values[n];

  for ( i = n+1; i < total ; i++)
  {
    b = values[i] - a;
    c = b*b;

    if( c < 10)
        newvalues[i] = c;
    }
}

This is what I have currently, but it does not seem to be giving the correct results? does anyone know what I'm doing wrong. Cheers

__global__ void calc(int total, float *values, float *newvalues){

float a,b,c;

int idx = blockIdx.x * blockDim.x + threadIdx.x;

for (int n = idx; n < (total-1); n += blockDim.x*gridDim.x){
    a = values[n];

    for(int i = n+1; i < total; i++){
        b = values[i] - a;
        c = b*b;

    if( c < 10)
        newvalues[i] = c;

    }
}
like image 865
Roger Avatar asked Mar 15 '11 00:03

Roger


1 Answers

Realize this problem in 2D and launch your kernel with 2D thread blocks. The total number of threads in x and y dimension will be equal to total . The kernel code should look like this:

__global__ void calc(float *values, float *newvalues, int total){


float a,b,c;

int n= blockIdy.y * blockDim.y + threadIdx.y;
int i= blockIdx.x * blockDim.x + threadIdx.x;

  if (n>=total || i>=total)
        return;

a = values[n];
b = values[i] - a;
c = b*b;
 if( c < 10)
        newvalues[i] = c;  

// I don't know your problem statement but i think it should be like: newvalues[n*total+i] = c;  


}

Update:

This is how you should call the kernel

dim3 block(16,16);
dim3 grid (  (total+15)/16,  (total+15)/16  );
calc<<<grid,block>>>(float *val, float *newval, int T);

Also make sure you add this line in kernel (see updated kernel)

if (n>=total || i>=total)
return;
like image 126
jwdmsd Avatar answered Nov 09 '22 01:11

jwdmsd