CUDA - multiple kernels to compute a single value

Question

Hey, I'm trying to write a kernel to essentially do the following in C

 float sum = 0.0;
 for(int i = 0; i < N; i++){
   sum += valueArray[i]*valueArray[i];
 }
 sum += sum / N;

At the moment I have this inside my kernel, but it is not giving correct values.

int i0 = blockIdx.x * blockDim.x + threadIdx.x;

   for(int i=i0; i<N; i += blockDim.x*gridDim.x){
        *d_sum += d_valueArray[i]*d_valueArray[i];
    }

  *d_sum= __fdividef(*d_sum, N);

The code used to call the kernel is

  kernelName<<<64,128>>>(N, d_valueArray, d_sum);
  cudaMemcpy(&sum, d_sum, sizeof(float) , cudaMemcpyDeviceToHost);

I think that each kernel is calculating a partial sum, but the final divide statement is not taking into account the accumulated value from each of the threads. Every kernel is producing it's own final value for d_sum?

Does anyone know how could I go about doing this in an efficient way? Maybe using shared memory between threads? I'm very new to GPU programming. Cheers

Answer 1

You're updating d_sum from multiple threads.

See the following SDK sample:

http://developer.download.nvidia.com/compute/cuda/sdk/website/samples.html

Here's the code from that sample. Note how it's a two step process. Sum each thread block and then __syncthreads before attempting to accumulate the final result.

#define ACCUM_N 1024
__global__ void scalarProdGPU(
    float *d_C,
    float *d_A,
    float *d_B,
    int vectorN,
    int elementN
){
    //Accumulators cache
    __shared__ float accumResult[ACCUM_N];

    ////////////////////////////////////////////////////////////////////////////
    // Cycle through every pair of vectors,
    // taking into account that vector counts can be different
    // from total number of thread blocks
    ////////////////////////////////////////////////////////////////////////////
    for(int vec = blockIdx.x; vec < vectorN; vec += gridDim.x){
        int vectorBase = IMUL(elementN, vec);
        int vectorEnd  = vectorBase + elementN;

        ////////////////////////////////////////////////////////////////////////
        // Each accumulator cycles through vectors with
        // stride equal to number of total number of accumulators ACCUM_N
        // At this stage ACCUM_N is only preferred be a multiple of warp size
        // to meet memory coalescing alignment constraints.
        ////////////////////////////////////////////////////////////////////////
        for(int iAccum = threadIdx.x; iAccum < ACCUM_N; iAccum += blockDim.x){
            float sum = 0;

            for(int pos = vectorBase + iAccum; pos < vectorEnd; pos += ACCUM_N)
                sum += d_A[pos] * d_B[pos];

            accumResult[iAccum] = sum;
        }

        ////////////////////////////////////////////////////////////////////////
        // Perform tree-like reduction of accumulators' results.
        // ACCUM_N has to be power of two at this stage
        ////////////////////////////////////////////////////////////////////////
        for(int stride = ACCUM_N / 2; stride > 0; stride >>= 1){
            __syncthreads();
            for(int iAccum = threadIdx.x; iAccum < stride; iAccum += blockDim.x)
                accumResult[iAccum] += accumResult[stride + iAccum];
        }

        if(threadIdx.x == 0) d_C[vec] = accumResult[0];
    }
}