c++ - CUDA图像处理错误

Question

我正在从事一个小型图像处理项目。我想运行一个执行图像减法的 CUDA 程序。所以你有图像背景和具有相同背景但上面有一些其他东西的图​​像。一旦你减去图像，你就会得到剩下的东西。两张图片都是480*360，我的gpu是GTX780。我的程序抛出错误 ./main': free(): invalid next size (normal): 0x000000000126bd70 ***
中止(核心转储) 并且输出图像错误。我一直在努力解决这个问题。这是代码:

内核:

__global__ void add(unsigned char* a, unsigned char* b, unsigned char* c, int numCols, int numWidth) {
    int i = blockIdx.x * blockDim.x + threadIdx.x; //Column
    int j = blockIdx.y * blockDim.y + threadIdx.y; //Row
    if(i < numWidth && j < numCols)
    {
      int idx = j * numCols + i;
      c[idx] = b[idx] - a[idx];
    }   
}

和主要功能:

int main() {
    CImg<unsigned char> img1("1.bmp");
    CImg<unsigned char> img2("2.bmp");
    //both images have the same size
    int width = img1.width();
    int height = img1.height();

    int size = width * height * 3; //both images of same size

    dim3 blockSize(16, 16, 1);
    dim3 gridSize((width + blockSize.x - 1) / blockSize.x, (height + blockSize.y - 1) / blockSize.y, 1);

    unsigned char *dev_a, *dev_b, *dev_c;

    cudaMalloc((void**)&dev_a, size * (sizeof(unsigned char)));
    cudaMalloc((void**)&dev_b, size * (sizeof(unsigned char)));
    cudaMalloc((void**)&dev_c, size * (sizeof(unsigned char)));

    cudaMemcpy(dev_a, img1, size * (sizeof(unsigned char)), cudaMemcpyHostToDevice);
    cudaMemcpy(dev_b, img2, size * (sizeof(unsigned char)), cudaMemcpyHostToDevice);

    add<<<gridSize, blockSize>>>(dev_a, dev_b, dev_c, height, width);

    cudaMemcpy(img2, dev_c, size * (sizeof(unsigned char)), cudaMemcpyDeviceToHost);

    img2.save("out.bmp");
    cudaFree(dev_a);
    cudaFree(dev_b);
    cudaFree(dev_c);
    return 0;
}

图像加载了 CImg图书馆。

Answer 1

问题在于在主机代码中错误地使用了 cimg 容器。根据documentation , 图像数据指针是通过data() 方法访问的，这意味着宿主代码中的cudaMemcpy 调用应该由img1.data()< 提供 和 img2.data()。

[此答案根据评论汇总并添加为社区 wiki 条目]