cuda - 在主机和 GPU 上添加 CUDA 的不同结果

Question

我有一个函数，它拍摄彩色图片并返回它的灰色版本。
如果我在主机上运行顺序代码，一切正常。如果我在设备上运行它，结果会略有不同(与正确值相比，1000 中的一个像素是 +1 或 -1)。

我认为这与转换有关，但我不确定。这是我使用的代码:

    __global__ void rgb2gray_d (unsigned char *deviceImage, unsigned char *deviceResult, const int height, const int width){
    /* calculate the global thread id*/
    int threadsPerBlock  = blockDim.x * blockDim.y;
    int threadNumInBlock = threadIdx.x + blockDim.x * threadIdx.y;
    int blockNumInGrid   = blockIdx.x  + gridDim.x  * blockIdx.y;

    int globalThreadNum = blockNumInGrid * threadsPerBlock + threadNumInBlock;
    int i = globalThreadNum;

    float grayPix = 0.0f;
    float r = static_cast< float >(deviceImage[i]);
    float g = static_cast< float >(deviceImage[(width * height) + i]);
    float b = static_cast< float >(deviceImage[(2 * width * height) + i]);
    grayPix = (0.3f * r) + (0.59f * g) + (0.11f * b);

    deviceResult[i] = static_cast< unsigned char > (grayPix);
}

void rgb2gray(unsigned char *inputImage, unsigned char *grayImage, const int width, const int height, NSTimer &timer) {

    unsigned char *deviceImage;
    unsigned char *deviceResult;

    int initialBytes = width * height * 3;  
    int endBytes =  width * height * sizeof(unsigned char);

    unsigned char grayImageSeq[endBytes];

    cudaMalloc((void**) &deviceImage, initialBytes);
    cudaMalloc((void**) &deviceResult, endBytes);
    cudaMemset(deviceResult, 0, endBytes);
    cudaMemset(deviceImage, 0, initialBytes);

    cudaError_t err = cudaMemcpy(deviceImage, inputImage, initialBytes, cudaMemcpyHostToDevice);    

    // Convert the input image to grayscale 
    rgb2gray_d<<<width * height / 256, 256>>>(deviceImage, deviceResult, height, width);
    cudaDeviceSynchronize();

    cudaMemcpy(grayImage, deviceResult, endBytes, cudaMemcpyDeviceToHost);

    ////// Sequential
    for ( int y = 0; y < height; y++ ) {
             for ( int x = 0; x < width; x++ ) {
                   float grayPix = 0.0f;
                   float r = static_cast< float >(inputImage[(y * width) + x]);
                   float g = static_cast< float >(inputImage[(width * height) + (y * width) + x]);
                   float b = static_cast< float >(inputImage[(2 * width * height) + (y * width) + x]);

                   grayPix = (0.3f * r) + (0.59f * g) + (0.11f * b);
                   grayImageSeq[(y * width) + x] = static_cast< unsigned char > (grayPix);
              }
        }

    //compare sequential and cuda and print pixels that are wrong
    for (int i = 0; i < endBytes; i++)
    {
        if (grayImage[i] != grayImageSeq[i])
        cout << i << "-" << static_cast< unsigned int >(grayImage[i]) <<
                 " should be " << static_cast< unsigned int >(grayImageSeq[i]) << endl;
        }

    cudaFree(deviceImage);
    cudaFree(deviceResult);
}

我提到我为初始图像分配了宽度 * 高度 * 3，因为初始图像是一个 CImg。

我在 GeForce GTX 480 上工作。

Answer 1

最后我找到了答案。 CUDA 会自动融合单精度和 double 的乘加。使用下面的文档 1 , 第 4.4 节，我设法修复了它。而不是做

grayPix = (0.3f * r) + (0.59f * g) + (0.11f * b);

我现在正在做

grayPix = __fadd_rn(__fadd_rn(__fmul_rn(0.3f, r),__fmul_rn(0.59f, g)), __fmul_rn(0.11f, b));

这将禁用乘法合并并添加到融合乘加指令中。

Floating Point and IEEE 754 Compliance for NVIDIA GPUs