CUDA 袖口 2D 示例

我目前正在开发一个必须实现2D-FFT的程序（用于互相关）。我用 CUDA 做了一个 1D FFT，它给了我正确的结果，我现在正在尝试实现一个 2D 版本。由于网上的例子和文档很少，我发现很难找出错误是什么。

到目前为止，我一直在使用 cuFFT 手册。

无论如何，我已经创建了两个 5x5 数组并用 1 填充它们。我已经将它们复制到 GPU 内存中并完成了前向 FFT，将它们相乘，然后在结果上完成 ifft。这给了我一个值为 650 的 5x5 数组。我希望在 5x5 阵列中的一个插槽中仅获得值为 25 的 DC 信号。相反，我在整个数组中得到 650。

此外，在将信号复制到 GPU 内存后，我不允许打印出信号的值。写作

cout << d_signal[1].x << endl;

给我一个访问违规。我在其他 cuda 程序中做了同样的事情，这不是问题。它与复变量的工作方式有关，还是人为错误？

如果有人有任何关于出错的指示，我将不胜感激。这是代码

   #include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <helper_functions.h>
#include <helper_cuda.h>
#include <ctime>
#include <time.h>
#include <stdio.h>
#include <iostream>
#include <math.h>
#include <cufft.h>
#include <fstream>
using namespace std;
typedef float2 Complex;


__global__ void ComplexMUL(Complex *a, Complex *b)
{
    int i = threadIdx.x;
    a[i].x = a[i].x * b[i].x - a[i].y*b[i].y;
    a[i].y = a[i].x * b[i].y + a[i].y*b[i].x;
}

int main()
{

    int N = 5;
    int SIZE = N*N;

    Complex *fg = new Complex[SIZE];
    for (int i = 0; i < SIZE; i++){
        fg[i].x = 1; 
        fg[i].y = 0;
    }
    Complex *fig = new Complex[SIZE];
    for (int i = 0; i < SIZE; i++){
        fig[i].x = 1; // 
        fig[i].y = 0;
    }
    for (int i = 0; i < 24; i=i+5)
    {
        cout << fg[i].x << " " << fg[i + 1].x << " " << fg[i + 2].x << " " << fg[i + 3].x << " " << fg[i + 4].x << endl;
    }
    cout << "----------------" << endl;
    for (int i = 0; i < 24; i = i + 5)
    {
        cout << fig[i].x << " " << fig[i + 1].x << " " << fig[i + 2].x << " " << fig[i + 3].x << " " << fig[i + 4].x << endl;
    }
    cout << "----------------" << endl;
    int mem_size = sizeof(Complex)* SIZE;

    cufftComplex *d_signal;
    checkCudaErrors(cudaMalloc((void **) &d_signal, mem_size)); 
    checkCudaErrors(cudaMemcpy(d_signal, fg, mem_size, cudaMemcpyHostToDevice));
    cufftComplex *d_filter_kernel;
    checkCudaErrors(cudaMalloc((void **)&d_filter_kernel, mem_size));
    checkCudaErrors(cudaMemcpy(d_filter_kernel, fig, mem_size, cudaMemcpyHostToDevice));
    // cout << d_signal[1].x << endl;
    // CUFFT plan
    cufftHandle plan;
    cufftPlan2d(&plan, N, N, CUFFT_C2C);
    // Transform signal and filter
    printf("Transforming signal cufftExecR2Cn");
    cufftExecC2C(plan, (cufftComplex *)d_signal, (cufftComplex *)d_signal, CUFFT_FORWARD);
    cufftExecC2C(plan, (cufftComplex *)d_filter_kernel, (cufftComplex *)d_filter_kernel, CUFFT_FORWARD);
    printf("Launching Complex multiplication<<< >>>n");
    ComplexMUL <<< 32, 256 >> >(d_signal, d_filter_kernel);
    // Transform signal back
    printf("Transforming signal back cufftExecC2Cn");
    cufftExecC2C(plan, (cufftComplex *)d_signal, (cufftComplex *)d_signal, CUFFT_INVERSE);
    Complex *result = new Complex[SIZE];
    cudaMemcpy(result, d_signal, sizeof(Complex)*SIZE, cudaMemcpyDeviceToHost);
    for (int i = 0; i < SIZE; i=i+5)
    {
        cout << result[i].x << " " << result[i + 1].x << " " << result[i + 2].x << " " << result[i + 3].x << " " << result[i + 4].x << endl;
    }
    delete result, fg, fig;
    cufftDestroy(plan);
    //cufftDestroy(plan2);
    cudaFree(d_signal);
    cudaFree(d_filter_kernel);
}

上面的代码给出了以下终端输出：

1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
----------------
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
----------------
Transforming signal cufftExecR2C
Launching Complex multiplication<<< >>>
Transforming signal back cufftExecC2C
625 625 625 625 625
625 625 625 625 625
625 625 625 625 625
625 625 625 625 625
625 625 625 625 625