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我正在做一个图像旋转方法。它需要两个矩阵和一定程度的旋转。它按度数旋转原始矩阵并将其存储到旋转矩阵中。我有以下“正常”代码（用于 CPU - 取自本网站 - http://sinepost.wordpress.com/2012/07/24/image-rotation/）并且它正在正常工作；

static void RotateImage(unsigned char original[RAW_HEIGHT][RAW_WIDTH] , unsigned char rotated[RAW_HEIGHT][RAW_WIDTH] , int degrees)
{
    double centerX = RAW_WIDTH/2;
    double centerY = RAW_HEIGHT/2;

    for(int x = 0; x< RAW_HEIGHT;x++)
    {
        for (int y = 0; y < RAW_WIDTH; y++)
        {
            double dir = calculateDirection(x-centerX,y-centerY);
            double mag = calculateMagnitude(x-centerX,y-centerY);

            dir-=degrees;

            int origX = (int)(centerX + calculateX(dir,mag));
            int origY = (int)(centerY + calculateY(dir,mag));

            if (origX >= 0 && origX < RAW_HEIGHT && origY >= 0 && origY < RAW_WIDTH)
            {
                    rotated[x][y] = original[origX][origY];
            }
        }
    }
}

我想将此代码转移到 CUDA 代码。这是我的版本：

#define RAW_WIDTH 1600*3
#define RAW_HEIGHT 1200

unsigned char *dev_original_image;
unsigned char *dev_rotated_image;

__global__ void rotatePicture(unsigned char *original, unsigned char *rotated, int degrees)
{
    int x = threadIdx.x + blockDim.x * blockIdx.x;
    int y = threadIdx.y + blockDim.y * blockIdx.y;
    int offset_rotated = x + y * blockDim.x * gridDim.x;

    double centerX = 2400.0;
    double centerY = 600.0;

    double dir = (atan2(y-centerY,x-centerX))*180/3.14159265;
    double mag = sqrt((x-centerX)*(x-centerX) + (y-centerY)*(y-centerY));

    dir = dir - degrees;

    int origX = (int)(centerX + cos((dir*3.14159265/180)) * mag);
    int origY = (int)(centerY + sin((dir*3.14159265/180)) * mag);
    int offset_original = origX + origY * blockDim.x * gridDim.x;

    if(offset_original > 0 && offset_original < RAW_HEIGHT*RAW_WIDTH)
        *(rotated + offset_rotated) = *(original + offset_original);
}

但它并没有给我与 CPU 部分相同的结果。我认为问题在于传递 CUDA kerenl 的参数。我将它们作为二维数组传递，可以吗？谁可以给我解释一下这个？这是我的 kerenl 配置和调用：

dim3 BlockPerGrid(450,400,1);
dim3 ThreadsPerGrid(8,4,1);

cudaMalloc((void**)&dev_original_image,sizeof(unsigned char)*RAW_HEIGHT*RAW_WIDTH);
cudaMalloc((void**)&dev_rotated_image,sizeof(unsigned char)*RAW_HEIGHT*RAW_WIDTH);

cudaMemcpy(dev_original_image, raw_image2D, sizeof(unsigned char)*RAW_HEIGHT*RAW_WIDTH,cudaMemcpyHostToDevice);
cudaMemcpy(dev_rotated_image, raw_image2D_rotated, sizeof(unsigned char)*RAW_HEIGHT*RAW_WIDTH, cudaMemcpyHostToDevice);

rotatePicture<<<BlockPerGrid,ThreadsPerGrid>>>(dev_original_image,dev_rotated_image, deg);

感谢您的建议！

注意：我修改了我的代码并且工作得更好但仍然不正确。

4

1 回答 1

1

这是潜伏在这些水域中的其他人的解决方案。这是我正确的内核：

__global__ void rotatePicture(unsigned char *original, unsigned char *rotated, int degrees)
{
    int x = threadIdx.x + blockDim.x * blockIdx.x;
    int y = threadIdx.y + blockDim.y * blockIdx.y;
    int offset_rotated = x + y * blockDim.x * gridDim.x;

    double centerX = 2400.0;
    double centerY = 600.0;

    double dir = (atan2(x-centerX,y-centerY))*180/3.14159265;
    double mag = sqrt((x-centerX)*(x-centerX) + (y-centerY)*(y-centerY));

    dir = dir - degrees;

    int origX = (int)(centerX + sin((dir*3.14159265/180)) * mag);
    int origY = (int)(centerY + cos((dir*3.14159265/180)) * mag);
    int offset_original = origX + origY * blockDim.x * gridDim.x;

    if(origX > 0 && origX < RAW_WIDTH && origY > 0 && origY < RAW_HEIGHT)
        *(rotated + offset_rotated) = *(original + offset_original);
}

另外，我像这样更改了内核尺寸（以适应我的 1600*3 宽度和 1200 高度）：

dim3 BlockPerGrid(600,300,1);
dim3 ThreadsPerGrid(8,4,1);

因此，它的运行方式与上述 CPU 版本相同，但使用的是 GPU 资源。享受

于 2013-02-18T22:04:53.350 回答