尽管代码中没有双打,但Cuda降低了双重浮动错误

时间:2011-11-29 04:14:08

标签: cuda gpu pycuda

我正在使用PyCUDA编写内核。我的GPU设备仅支持计算能力1.1(arch sm_11),因此我只能在代码中使用浮点数。我已经花了很大力气确保我用浮点数做所有事情,但尽管如此,我的代码中还有一条特殊的行会导致编译错误。

代码块是:

  // Gradient magnitude, so 1 <= x <= width, 1 <= y <= height. 
  if( j > 0 && j < im_width && i > 0 && i < im_height){
    gradient_mag[idx(i,j)] = float(sqrt(x_gradient[idx(i,j)]*x_gradient[idx(i,j)] + y_gradient[idx(i,j)]*y_gradient[idx(i,j)]));
  }

这里,idx()是一个__device__辅助函数,它根据像素索引ij返回一个线性索引,它只适用于整数。我一直在使用它,它不会在其他任何地方给出错误,所以我强烈怀疑它不是idx()sqrt()调用仅来自支持浮点数的标准C数学函数。所涉及的所有数组x_gradienty_gradientgradient_mag都是float*,它们是我函数输入的一部分(即在Python中声明,然后转换为设备变量等。)。

我已经尝试将额外的强制转换移到上面的代码中,没有运气。我也尝试过像这样完全愚蠢的事情:

 // Gradient magnitude, so 1 <= x <= width, 1 <= y <= height. 
 if( j > 0 && j < im_width && i > 0 && i < im_height){
    gradient_mag[idx(i,j)] = 3.0f; // also tried float(3.0) here
  }

所有这些变化都会产生同样的错误:

 pycuda.driver.CompileError: nvcc said it demoted types in source code it compiled--this is likely not what you want.
 [command: nvcc --cubin -arch sm_11 -I/usr/local/lib/python2.7/dist-packages/pycuda-2011.1.2-py2.7-linux-x86_64.egg/pycuda/../include/pycuda kernel.cu]
 [stderr:
 ptxas /tmp/tmpxft_00004329_00000000-2_kernel.ptx, line 128; warning : Double is not supported. Demoting to float
 ]

有什么想法吗?我在我的代码中调试了很多错误,希望今晚能让它工作,但事实证明这是一个我无法理解的错误。

已添加 - 这是内核的截断版本,在我的计算机上产生相同的错误。

 every_pixel_hog_kernel_source = \
 """
 #include <math.h>
 #include <stdio.h>

 __device__ int idx(int ii, int jj){
     return gridDim.x*blockDim.x*ii+jj;
 }

 __device__ int bin_number(float angle_val, int total_angles, int num_bins){ 

     float angle1;   
     float min_dist;
     float this_dist;
     int bin_indx;

     angle1 = 0.0;
     min_dist = abs(angle_val - angle1);
     bin_indx = 0;

     for(int kk=1; kk < num_bins; kk++){
         angle1 = angle1 + float(total_angles)/float(num_bins);
         this_dist = abs(angle_val - angle1);
         if(this_dist < min_dist){
             min_dist = this_dist;
             bin_indx = kk;
         }
     }

     return bin_indx;
 }

 __device__ int hist_number(int ii, int jj){

     int hist_num = 0;

     if(jj >= 0 && jj < 11){ 
         if(ii >= 0 && ii < 11){ 
             hist_num = 0;
         }
         else if(ii >= 11 && ii < 22){
             hist_num = 3;
         }
         else if(ii >= 22 && ii < 33){
             hist_num = 6;
         }
     }
     else if(jj >= 11 && jj < 22){
         if(ii >= 0 && ii < 11){ 
             hist_num = 1;
         }
         else if(ii >= 11 && ii < 22){
             hist_num = 4;
         }
         else if(ii >= 22 && ii < 33){
             hist_num = 7;
         }
     }
     else if(jj >= 22 && jj < 33){
         if(ii >= 0 && ii < 11){ 
             hist_num = 2;
         }
         else if(ii >= 11 && ii < 22){
             hist_num = 5;
         }
         else if(ii >= 22 && ii < 33){
             hist_num = 8;
         }
     }

     return hist_num;
 }

  __global__ void every_pixel_hog_kernel(float* input_image, int im_width, int im_height, float* gaussian_array, float* x_gradient, float* y_gradient, float* gradient_mag, float* angles, float* output_array)
  {    
      /////
      // Setup the thread indices and linear offset.
      /////
      int i = blockDim.y * blockIdx.y + threadIdx.y;
      int j = blockDim.x * blockIdx.x + threadIdx.x;
      int ang_limit = 180;
      int ang_bins = 9;
      float pi_val = 3.141592653589f; //91

      /////
      // Compute a Gaussian smoothing of the current pixel and save it into a new image array
      // Use sync threads to make sure everyone does the Gaussian smoothing before moving on.
      /////
      if( j > 1 && i > 1 && j < im_width-2 && i < im_height-2 ){

            // Hard-coded unit standard deviation 5-by-5 Gaussian smoothing filter.
            gaussian_array[idx(i,j)] = float(1.0/273.0) *(
            input_image[idx(i-2,j-2)] + float(4.0)*input_image[idx(i-2,j-1)] + float(7.0)*input_image[idx(i-2,j)] + float(4.0)*input_image[idx(i-2,j+1)] + input_image[idx(i-2,j+2)] + 
            float(4.0)*input_image[idx(i-1,j-2)] + float(16.0)*input_image[idx(i-1,j-1)] + float(26.0)*input_image[idx(i-1,j)] + float(16.0)*input_image[idx(i-1,j+1)] + float(4.0)*input_image[idx(i-1,j+2)] +
            float(7.0)*input_image[idx(i,j-2)] + float(26.0)*input_image[idx(i,j-1)] + float(41.0)*input_image[idx(i,j)] + float(26.0)*input_image[idx(i,j+1)] + float(7.0)*input_image[idx(i,j+2)] +
            float(4.0)*input_image[idx(i+1,j-2)] + float(16.0)*input_image[idx(i+1,j-1)] + float(26.0)*input_image[idx(i+1,j)] + float(16.0)*input_image[idx(i+1,j+1)] + float(4.0)*input_image[idx(i+1,j+2)] +
            input_image[idx(i+2,j-2)] + float(4.0)*input_image[idx(i+2,j-1)] + float(7.0)*input_image[idx(i+2,j)] + float(4.0)*input_image[idx(i+2,j+1)] + input_image[idx(i+2,j+2)]);
     }
     __syncthreads();

     /////
     // Compute the simple x and y gradients of the image and store these into new images
     // again using syncthreads before moving on.
     /////

     // X-gradient, ensure x is between 1 and width-1
     if( j > 0 && j < im_width){
         x_gradient[idx(i,j)] = float(input_image[idx(i,j)] - input_image[idx(i,j-1)]);
     }
     else if(j == 0){
         x_gradient[idx(i,j)] = float(0.0);
     }

    // Y-gradient, ensure y is between 1 and height-1
    if( i > 0 && i < im_height){
         y_gradient[idx(i,j)] = float(input_image[idx(i,j)] - input_image[idx(i-1,j)]);
    }
    else if(i == 0){
        y_gradient[idx(i,j)] = float(0.0);
    }
    __syncthreads();

    // Gradient magnitude, so 1 <= x <= width, 1 <= y <= height. 
    if( j < im_width && i < im_height){

        gradient_mag[idx(i,j)] = float(sqrt(x_gradient[idx(i,j)]*x_gradient[idx(i,j)] + y_gradient[idx(i,j)]*y_gradient[idx(i,j)]));
    }
    __syncthreads();

    /////
    // Compute the orientation angles
    /////
    if( j < im_width && i < im_height){
        if(ang_limit == 360){
            angles[idx(i,j)] = float((atan2(y_gradient[idx(i,j)],x_gradient[idx(i,j)])+pi_val)*float(180.0)/pi_val);
        }
        else{
            angles[idx(i,j)] = float((atan( y_gradient[idx(i,j)]/x_gradient[idx(i,j)] )+(pi_val/float(2.0)))*float(180.0)/pi_val);
        }
    }
    __syncthreads();

    // Compute the HoG using the above arrays. Do so in a 3x3 grid, with 9 angle bins for each grid.
    // forming an 81-vector and then write this 81 vector as a row in the large output array.

    int top_bound, bot_bound, left_bound, right_bound, offset;
    int window = 32;

    if(i-window/2 > 0){
        top_bound = i-window/2;
        bot_bound = top_bound + window;
    }
    else{
        top_bound = 0;
        bot_bound = top_bound + window;
    }

    if(j-window/2 > 0){
        left_bound = j-window/2;
        right_bound = left_bound + window;
    }
    else{
        left_bound = 0;
        right_bound = left_bound + window;
    }

    if(bot_bound - im_height > 0){
        offset = bot_bound - im_height;
        top_bound = top_bound - offset;
        bot_bound = bot_bound - offset;
    }

    if(right_bound - im_width > 0){
        offset = right_bound - im_width;
        right_bound = right_bound - offset;
        left_bound = left_bound - offset;
    }

    int counter_i = 0;
    int counter_j = 0;
    int bin_indx, hist_indx, glob_col_indx, glob_row_indx;
    int row_width = 81; 

    for(int pix_i = top_bound; pix_i < bot_bound; pix_i++){
        for(int pix_j = left_bound; pix_j < right_bound; pix_j++){

            bin_indx = bin_number(angles[idx(pix_i,pix_j)], ang_limit, ang_bins);
            hist_indx = hist_number(counter_i,counter_j);

            glob_col_indx = ang_bins*hist_indx + bin_indx;
            glob_row_indx = idx(i,j);

            output_array[glob_row_indx*row_width + glob_col_indx] = float(output_array[glob_row_indx*row_width + glob_col_indx] + float(gradient_mag[idx(pix_i,pix_j)]));


            counter_j = counter_j + 1; 
        }
        counter_i = counter_i + 1;
        counter_j = 0;
    }

}
"""

3 个答案:

答案 0 :(得分:3)

这是一个使用双打的明显例子:

 gaussian_array[idx(i,j)] = float(1.0/273.0) *

看到正在划分的双重文字?

但实际上,使用float文字而不是双重文字强制转换为浮点数 - 强制转换是非常丑陋的,我建议他们隐藏这样的错误。

-------编辑1 / Dec ---------

首先,感谢@ CygnusX1,常量折叠会阻止这种计算 - 我甚至没想到它。

我试图重现错误的环境:我安装了CUDA SDK 3.2(@EMS已经提到它们似乎在实验室中使用),编译上面的截断内核版本,实际上nvopencc确实优化了上面的内容计算结束(感谢@ CygnusX1),实际上它并没有在生成的PTX代码中的任何地方使用双精度数。此外,ptxas没有给出@EMS收到的错误。从那以后,我认为这个问题不在every_pixel_hog_kernel_source代码本身之内,也许在PyCUDA中。但是,使用PyCUDA 2011.1.2并使用仍然进行编译不会产生类似@ EMS的问题的警告。我可以在问题中得到错误,但是通过引入双重计算,例如从gaussian_array[idx(i,j)] = float(1.0/273.0) *中移除强制转换

要获得相同的python案例,以下是否会产生错误:

import pycuda.driver as cuda
from pycuda.compiler import compile

x=compile("""put your truncated kernel code here""",options=[],arch="sm_11",keep=True)

在我的环境中不会产生错误,因此我可能无法复制您的结果。 不过,我可以提一些建议。使用compile(或SourceModule)时,如果使用keep=True,python将在显示错误消息之前打印出生成ptx文件的文件夹。 然后,如果你可以检查在该文件夹中生成的ptx文件,并查看.f64出现在哪里,它应该知道什么被视为双 - 但是,解密原始内核中的代码是困难的 - 拥有产生错误的最简单示例将对您有所帮助。

答案 1 :(得分:1)

你的问题在这里:

angle1 = 0.0;

0.0是双精度常数。 0.0f是单精度常数。

答案 2 :(得分:0)

(评论,不是答案,但它太大了,不能作为评论)

您能否在发生错误的行周围提供PTX代码?

我尝试使用您提供的代码编译一个简单的内核:

__constant__ int im_width;
__constant__ int im_height;

__device__ int idx(int i,int j) {
    return i+j*im_width;
}

__global__ void kernel(float* gradient_mag, float* x_gradient, float* y_gradient) {
    int i = threadIdx.x;
    int j = threadIdx.y;
  // Gradient magnitude, so 1 <= x <= width, 1 <= y <= height.
  if( j > 0 && j < im_width && i > 0 && i < im_height){
    gradient_mag[idx(i,j)] = float(sqrt(x_gradient[idx(i,j)]*x_gradient[idx(i,j)] + y_gradient[idx(i,j)]*y_gradient[idx(i,j)]));
  }
}

使用:

nvcc.exe -m32 -maxrregcount = 32 -gencode = arch = compute_11,code = \“sm_11,compute_11 \”--compile -o“Debug \ main.cu.obj”main.cu

没有错误。

使用CUDA 4.1 beta编译器


<强>更新

我尝试编译你的新代码(我在CUDA / C ++中工作,而不是PyCUDA,但这不重要)。没有抓到错误!使用CUDA 4.1和CUDA 4.0。 您的CUDA安装版本是什么?

C:\>nvcc --version
nvcc: NVIDIA (R) Cuda compiler driver
Copyright (c) 2005-2011 NVIDIA Corporation
Built on Wed_Oct_19_23:13:02_PDT_2011
Cuda compilation tools, release 4.1, V0.2.1221