将整数施放到Opencl中的浮子

Question

这是我在Stack Overflow上的第一篇文章，所以请忍受。

我目前正在编程OPENCL内核，并需要使用Innoluilt SQRT函数。但是，为此，该功能的参数必须是浮点。我目前有一个整数值，需要将其转换为float，以执行SQRT（）函数，然后将其转换回整数，以便将其存储到“ Magout”数组中。

下面的代码应该更好地了解我要做的事情：

magOutput[workItemNum] = sqrt(xConv[workItemNum]*xConv[workItemNum] + yConv[workItemNum]*yConv[workItemNum]);

如果需要了解所需的应用程序是完整的代码：

__kernel void matrixMultiplication(__global int* input, __global int* xConv, __global int* yConv, __global int* size, __global int* magOutput){

int workItemNum = get_global_id(0); //Work item ID
int workGroupNum = get_group_id(0); //Work group ID
int localGroupID = get_local_id(0); //Work items ID within each work group

// size refers to the total size of a matrix. So for a 3x3 size = 9
float dim = *size;
int dim1 = *size; 

int row = sqrt(dim); // only square matrices are used and as such the sqrt of size produces the row length
int current_row = workItemNum/dim; // the current row is calculated by using the current workitem number divided by the total size of the matrix

int col = sqrt(dim); // only square matrices are used and as such the sqrt of size produces the column length
int current_col = workItemNum % dim1; // the current column is calculated by using the current workitem number modulus by the total size of the matrix

// printf("dimension: %i \n",localGroupID);

// This if statement excludes all boundary pixels from the calculation as you require the neighbouring pixel cells 
// for this calculation
if (current_col == 0 || current_col == col-1 || current_row == 0 || current_row == row - 1){
    /*===============================================================================================================
    * The xConv array performs the kernal convultion of the input grey scale values with the following matrix:
    *
    *                            [-1  0 +1]
    * X - Directional Kernel  =  [-2  0 +2]
    *                            [-1  0 +1]
    * 
    * This scans across the X direction of the image and enhances all edges in the X-direction 
    * ===============================================================================================================
    */
    xConv[workItemNum] =  input[(current_col - 1)*col + current_row - 1]*-1 
             + input[(current_col)*col + current_row - 1]*0 
             + input[(current_col + 1)*col + current_row - 1]*1 
             + input[(current_col - 1)*col + current_row]*-2 
             + input[(current_col)*col + current_row]*0 
             + input[(current_col + 1)*col + current_row]*2 
             + input[(current_col - 1)*col + current_row + 1]*-1 
             + input[(current_col)*col + current_row + 1]*0 
             + input[(current_col + 1)*col + current_row + 1]*1;

    /*===============================================================================================================
    * The xConv array performs the kernal convultion of the input grey scale values with the following matrix:
    *
    *                            [+1 +2 +1]
    * Y - Directional Kernel  =  [ 0  0  0]
    *                            [-1 -2 -1]
    * 
    * This scans across the Y direction of the image and enhances all edges in the Y-direction 
    * ===============================================================================================================
    */
    yConv[workItemNum] = input[(current_col - 1)*col + current_row - 1]*-1 
             + input[(current_col)*col + current_row - 1]*-2 
             + input[(current_col + 1)*col + current_row - 1]*-1 
             + input[(current_col - 1)*col + current_row]*0 
             + input[(current_col)*col + current_row]*0 
             + input[(current_col + 1)*col + current_row]*0 
             + input[(current_col - 1)*col + current_row + 1]*1 
             + input[(current_col)*col + current_row + 1]*2 
             + input[(current_col + 1)*col + current_row + 1]*1;
}

//===============================================================================================================
// Calculates the convolution matrix of the X and Y arrays. Does so by squaring each item of the X and Y arrays,  
// adding them and taking the square root. This is the basic magnitude formula. This is done for by each workItem
//===============================================================================================================
magOutput[workItemNum] = sqrt(xConv[workItemNum]*xConv[workItemNum] + yConv[workItemNum]*yConv[workItemNum]);
}

是否有建议？

Answer 1

因此，您本质上拥有int数据类型的2D向量，并希望计算其长度。 OPENCL C的大部分仅是标准C99代码/语法，因此最直接的方法是使用标准的C风格类型铸造：

magOutput[workItemNum] = (int)(sqrt((float)xConv[workItemNum]*(float)xConv[workItemNum] + (float)yConv[workItemNum]*(float)yConv[workItemNum])+0.5f);

+0.5F用于正确的舍入：铸造float到int始终向下滚动，例如（int）3.9f将转换为3。通过在铸造之前立即添加+0.5F，将结果正确/向上舍入/下。请注意，我首先施放float，然后进行平方；否则，在乘法过程中可能会出现整数溢出。

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一个可能更快的方法是：在这里，我加载值xconv [workitemnum]/yconv [workitemnum]从全局内存中仅一次（这真的很慢），请施放它们到float并将它们存储在私有内存（寄存器）xconvf/yconvf。然后，我进行长度计算和舍入，然后将结果写回magOutput [workitemnum]在缓慢的全局内存中。

const float xConvf = (float)xConv[workItemNum], yConvf = (float)yConv[workItemNum];
magOutput[workItemNum] = (int)(sqrt(xConvf*xConvf + yConvf*yConvf)+0.5f);

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如果您想在OpenCl C中使用内置的数学功能非常喜欢，也可以执行此操作（应该与第二种方法一样快）：

magOutput[workItemNum] = (int)(length(float2((float)xConv[workItemNum], (float)yConv[workItemNum]))+0.5f);

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因此，您有2个要点：

• type in opencl中的casting opencl中的作品就像在C：（float）x中。
• 从 /写入到全局内存（您的内核参数阵列）确实很慢。仅在局部变量（这些是私人内存空间）中加载必要的值，以在寄存器中使用数字，然后在寄存器中进行算术，然后再回到全局内存中。

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虽然这使您的OpenCl C代码未触及，但我建议使用此轻量级 opencl-wrapper 用于开发C ++ 。这将CPU代码中的OPENCL控制逻辑降低到约1/4，并使开发变得更加容易。