Asynchronous executions of CUDA memory copies and

Question

I have a CUDA program for calculating FFTs of, let's say, size 50000. Currently, I copy the whole array to the GPU and execute the cuFFT. Now, I am trying to optimize the programm and the NVIDIA Visual Profiler tells me to hide the memcopy by concurrency with parallel computations. My question is:

Is it possible, for example, to copy the first 5000 Elements, then start calculating, then copying the next bunch of data in parallel to calculations etc?

Since a DFT is basically a sum over the time values multiplied with a complex exponential function, I think that it should possible to calculate the FFT "blockwise".

Does cufft support this? Is it in general a good computational idea?

EDIT

To be more clear, I do not want to calculate different FFTs parallel on different arrays. Lets say I have a big trace of a sinusoidal signal in the time domain and I want to know which frequencies are in the signal. My Idea is to copy, for example, one third of the signal length to the GPU, then the next third and calculate the FFT with the first third of the already copied input values parallel. Then copy the last third and update the output values until all the time values are processed. So in the end there should be one output array with a peak at the frequency of the sinus.

Answer 1

Please, take into account the comments above and, in particular, that:

1. If you calculate the FFT over Npartial elements, you will have an output of Npartial elements;
2. (following Robert Crovella) All the data required for the cuFFT must be resident on the device, before the cuFFT call is launched, so that you will not be able to break the data into pieces for a single cuFFT operation, and begin that operation before all pieces are on the GPU; furthermore, a cuFFT call is opaque;

Taking into account the above two points, I think you can only "emulate" what you like to achieve if you properly use zero padding in the way illustrated by the code below. As you will see, letting N to be the data size, by dividing the data in NUM_STREAMS chunks, the code performs NUM_STREAMS zero padded and streamed cuFFT calls of size N. After the cuFFT, you have to combine (sum) the partial results.

#include <stdio.h>

#include <cufft.h>

#define BLOCKSIZE 32
#define NUM_STREAMS 3

/**********/
/* iDivUp */
/*********/
int iDivUp(int a, int b) { return ((a % b) != 0) ? (a / b + 1) : (a / b); }

/********************/
/* CUDA ERROR CHECK */
/********************/
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
    if (code != cudaSuccess) 
    {
        fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
        if (abort) exit(code);
    }
}

/******************/
/* SUMMING KERNEL */
/******************/
__global__ void kernel(float2 *vec1, float2 *vec2, float2 *vec3, float2 *out, int N) {

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

    if (tid < N) {
        out[tid].x = vec1[tid].x + vec2[tid].x + vec3[tid].x;
        out[tid].y = vec1[tid].y + vec2[tid].y + vec3[tid].y;
    }

}


/********/
/* MAIN */
/********/
int main()
{
    const int N = 600000;
    const int Npartial = N / NUM_STREAMS;

    // --- Host input data initialization
    float2 *h_in1 = new float2[Npartial];
    float2 *h_in2 = new float2[Npartial];
    float2 *h_in3 = new float2[Npartial];
    for (int i = 0; i < Npartial; i++) {
        h_in1[i].x = 1.f;
        h_in1[i].y = 0.f;
        h_in2[i].x = 1.f;
        h_in2[i].y = 0.f;
        h_in3[i].x = 1.f;
        h_in3[i].y = 0.f;
    }

    // --- Host output data initialization
    float2 *h_out = new float2[N];

    // --- Registers host memory as page-locked (required for asynch cudaMemcpyAsync)
    gpuErrchk(cudaHostRegister(h_in1, Npartial*sizeof(float2), cudaHostRegisterPortable));
    gpuErrchk(cudaHostRegister(h_in2, Npartial*sizeof(float2), cudaHostRegisterPortable));
    gpuErrchk(cudaHostRegister(h_in3, Npartial*sizeof(float2), cudaHostRegisterPortable));

    // --- Device input data allocation
    float2 *d_in1;          gpuErrchk(cudaMalloc((void**)&d_in1, N*sizeof(float2)));
    float2 *d_in2;          gpuErrchk(cudaMalloc((void**)&d_in2, N*sizeof(float2)));
    float2 *d_in3;          gpuErrchk(cudaMalloc((void**)&d_in3, N*sizeof(float2)));
    float2 *d_out1;         gpuErrchk(cudaMalloc((void**)&d_out1, N*sizeof(float2)));
    float2 *d_out2;         gpuErrchk(cudaMalloc((void**)&d_out2, N*sizeof(float2)));
    float2 *d_out3;         gpuErrchk(cudaMalloc((void**)&d_out3, N*sizeof(float2)));
    float2 *d_out;          gpuErrchk(cudaMalloc((void**)&d_out, N*sizeof(float2)));

    // --- Zero padding
    gpuErrchk(cudaMemset(d_in1, 0, N*sizeof(float2)));
    gpuErrchk(cudaMemset(d_in2, 0, N*sizeof(float2)));
    gpuErrchk(cudaMemset(d_in3, 0, N*sizeof(float2)));

    // --- Creates CUDA streams
    cudaStream_t streams[NUM_STREAMS];
    for (int i = 0; i < NUM_STREAMS; i++) gpuErrchk(cudaStreamCreate(&streams[i]));

    // --- Creates cuFFT plans and sets them in streams
    cufftHandle* plans = (cufftHandle*) malloc(sizeof(cufftHandle)*NUM_STREAMS);
    for (int i = 0; i < NUM_STREAMS; i++) {
        cufftPlan1d(&plans[i], N, CUFFT_C2C, 1);
        cufftSetStream(plans[i], streams[i]);
    }

    // --- Async memcopyes and computations
    gpuErrchk(cudaMemcpyAsync(d_in1, h_in1, Npartial*sizeof(float2), cudaMemcpyHostToDevice, streams[0]));
    gpuErrchk(cudaMemcpyAsync(&d_in2[Npartial], h_in2, Npartial*sizeof(float2), cudaMemcpyHostToDevice, streams[1]));
    gpuErrchk(cudaMemcpyAsync(&d_in3[2*Npartial], h_in3, Npartial*sizeof(float2), cudaMemcpyHostToDevice, streams[2]));
    cufftExecC2C(plans[0], (cufftComplex*)d_in1, (cufftComplex*)d_out1, CUFFT_FORWARD);
    cufftExecC2C(plans[1], (cufftComplex*)d_in2, (cufftComplex*)d_out2, CUFFT_FORWARD);
    cufftExecC2C(plans[2], (cufftComplex*)d_in3, (cufftComplex*)d_out3, CUFFT_FORWARD);

    for(int i = 0; i < NUM_STREAMS; i++) gpuErrchk(cudaStreamSynchronize(streams[i]));

    kernel<<<iDivUp(BLOCKSIZE,N), BLOCKSIZE>>>(d_out1, d_out2, d_out3, d_out, N);
    gpuErrchk(cudaPeekAtLastError());
    gpuErrchk(cudaDeviceSynchronize());

    gpuErrchk(cudaMemcpy(h_out, d_out, N*sizeof(float2), cudaMemcpyDeviceToHost));

    for (int i=0; i<N; i++) printf("i = %i; real(h_out) = %f; imag(h_out) = %f\n", i, h_out[i].x, h_out[i].y);

    // --- Releases resources
    gpuErrchk(cudaHostUnregister(h_in1));
    gpuErrchk(cudaHostUnregister(h_in2));
    gpuErrchk(cudaHostUnregister(h_in3));
    gpuErrchk(cudaFree(d_in1));
    gpuErrchk(cudaFree(d_in2));
    gpuErrchk(cudaFree(d_in3));
    gpuErrchk(cudaFree(d_out1));
    gpuErrchk(cudaFree(d_out2));
    gpuErrchk(cudaFree(d_out3));
    gpuErrchk(cudaFree(d_out));

    for(int i = 0; i < NUM_STREAMS; i++) gpuErrchk(cudaStreamDestroy(streams[i]));

    delete[] h_in1;
    delete[] h_in2;
    delete[] h_in3;
    delete[] h_out;

    cudaDeviceReset();  

    return 0;
}

This is the timeline of the above code when run on a Kepler K20c card. As you can see, the computation overlaps the async memory transfers.