递归fft计算的分段错误

Das*_*rol 2 c++ math fft

我正在使用dif radix-2算法在复杂向量上执行fft.我递归地将我的输入分成2组,第一和第二半,然后为前半部分和后半部分执行复杂的加法*旋转因子.

函数完成但我在尝试输出结果向量时收到Segmentation错误.怎么了?

int main(int argc, char *argv[]){
    int n = 8;
    complex<double> *x = new complex<double>[n];

    // Test data
    x[0] = sin(M_PI/2);
    x[1] = sin(0);
    x[2] = sin(0);
    x[3] = sin(0);
    x[4] = sin(0);
    x[5] = sin(0);
    x[6] = sin(0);
    x[7] = sin(0);


    for(int i = 0; i<n; i++){
        cout << x[i] << endl;
    }

    fft(x,n);

    cout << endl;
    for(int i = 0; i<n; i++){
        cout << x[i] << endl;
    }

}
void fft(complex<double> *X, int N){
    if(N < 1){return;}

    double w = 2 * M_PI / (N/2);  

    for(int i = 0; i<N/2; i++){
        double ang = w * i;
        complex<double> tw(cos(ang),sin(ang));  // twiddle factor

        complex<double> first_half = X[i];
        complex<double> second_half = X[i+N/2];

        X[i] = first_half+second_half;
        X[i+N/2] = (first_half-second_half) * tw;

        cout << X[i] << " " <<X[i+N/2] << endl;;

    }
    fft(X,N-1);
    fft(X+N/2,N-1);
}
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Mik*_*our 7

fft(X+N/2,N-1);
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这将超出界限; 数组后半部分的大小只是N/2.我的傅里叶理论有点生疏,但我想你想要

fft(X, N/2);
fft(X+N/2, N/2);
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  • @DyP:那与'N/2`相同; 经典的FFT要求整体尺寸为2的幂,因此每个较小变换的尺寸始终是均匀的. (2认同)