c++ - 尝试计算 2 个图像的 FFT（快速傅里叶变换）时访问冲突

Question

我正在尝试对两个原始图像进行 FFT。但我得到了unhandled exception (access violation)- 我不知道为什么。我正在使用 fftw 库。首先我正在阅读两个图像，然后我计算 FFT。但在它开始计算之前，它会产生访问冲突错误。

#include "stdafx.h"
#include <iostream>
#include <conio.h>
#include "fftw3.h"

#define Width 2280
#define Height 170

unsigned char im2[170*2280];
unsigned char im1[170*2280];
float image1[170*2280];
float image2[170*2280];

using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{

    FILE* fp1, *fp2;

    //Read two images

    fp1 = fopen ("image1.raw" , "r");
    fread(im1, sizeof(unsigned char), Width* Height, fp1);

    fp2 = fopen ("image2.raw" , "r");
    fread(im2, sizeof(unsigned char), Width* Height, fp2);


    fclose(fp2);
    fclose(fp1);

    //Typecasting two images into float

    for (int i = 0; i < Width* Height; i++)
    {       
        image1[i]= (float)im1[i];
        image2[i] = (float)im2[i];

    }

    fftwf_plan fplan1, fplan2;
    fftwf_complex fft1[((Width/2)+1)*2];
    fftwf_complex fft2[((Width/2)+1)*2];

    fplan1 = fftwf_plan_dft_r2c_2d(Height, Width, (float*)image1, fft1, FFTW_ESTIMATE);
    fftwf_execute(fplan1);
    fftwf_destroy_plan(fplan1);

    fplan2 = fftwf_plan_dft_r2c_2d(Height,Width, image2, (fftwf_complex*)fft2, FFTW_ESTIMATE);
    fftwf_execute(fplan2);
    fftwf_destroy_plan(fplan2);

    _getch();
    return 0;
}

Answer 1

fft1 and fft2 are only large enough to hold one output row - you need Height rows. You'll probably want to allocate them dynamically too, as they will most likely be too large for the stack once you have the correct size, e.g.

fftwf_complex *fft1 = new fftwf_complex[((Width/2)+1)*2*Height];
fftwf_complex *fft2 = new fftwf_complex[((Width/2)+1)*2*Height];

NB: don't forget to call delete [] to free these later.