Question

我试图优化平滑函数，给定图像通过用图像周围的像素的平均值替换图像中的每个像素来平滑/模糊边缘（图像的尺寸是矩阵）。要优化的代码如下：

int i, j, ii, jj;
pixel_sum ps;

for (j = 0; j < dim; j++){
    for (i = 0; i < dim; i++){
        initialize_pixel_sum(&ps);
        for(ii = max(i-1, 0); ii <= min(i+1, dim-1); ii++){
            for(jj = max(j-1, 0); jj <= min(j+1, dim-1); jj++){
                accumulate_sum(&ps, src[RIDX(ii,jj,dim)]);
            }
        }
        dst[RIDX(i,j,dim)].red   = ps.red/ps.num;
        dst[RIDX(i,j,dim)].green = ps.green/ps.num;
        dst[RIDX(i,j,dim)].blue  = ps.blue/ps.num;
    }
}

我找到了一个优化版本，如下所示：

int i, j, myJ;

//cornors
dst[0].red = (src[0].red+src[1].red+src[dim].red+src[dim+1].red)>>2;
dst[0].blue = (src[0].blue+src[1].blue+src[dim].blue+src[dim+1].blue)>>2;
dst[0].green = (src[0].green+src[1].green+src[dim].green+src[dim+1].green)>>2;

i = dim*2-1;
dst[dim-1].red = (src[dim-2].red+src[dim-1].red+src[i-1].red+src[i].red)>>2;
dst[dim-1].blue = (src[dim-2].blue+src[dim-1].blue+src[i-1].blue+src[i].blue)>>2;
dst[dim-1].green = (src[dim-2].green+src[dim-1].green+src[i-1].green+src[i].green)>>2;

j = dim*(dim-1);
i = dim*(dim-2);
dst[j].red = (src[j].red+src[j + 1].red+src[i].red+src[i + 1].red)>>2;
dst[j].blue = (src[j].blue+src[j + 1].blue+src[i].blue+src[i + 1].blue)>>2;
dst[j].green = (src[j].green+src[j + 1].green+src[i].green+src[i + 1].green)>>2;

j = dim*dim-1;
i = dim*(dim-1)-1;
dst[j].red = (src[j - 1].red+src[j].red+src[i - 1].red+src[i].red)>>2;
dst[j].blue = (src[j - 1].blue+src[j].blue+src[i - 1].blue+src[i].blue)>>2;
dst[j].green = (src[j - 1].green+src[j].green+src[i - 1].green+src[i].green)>>2;

//sides
i = dim - 1;
for (j = 1; j < i; j++) 
{
    dst[j].red = (src[j].red+src[j-1].red+src[j+1].red+src[j+dim].red+src[j+1+dim].red+src[j-1+dim].red)/6;
    dst[j].green = (src[j].green+src[j-1].green+src[j+1].green+src[j+dim].green+src[j+1+dim].green+src[j-1+dim].green)/6;
    dst[j].blue = (src[j].blue+src[j-1].blue+src[j+1].blue+src[j+dim].blue+src[j+1+dim].blue+src[j-1+dim].blue)/6;
}

i = dim*dim-1;
for (j = i - dim + 2; j < i; j++) 
{
    dst[j].red = (src[j].red+src[j-1].red+src[j+1].red+src[j-dim].red+src[j+1-dim].red+src[j-1-dim].red)/6;
    dst[j].green = (src[j].green+src[j-1].green+src[j+1].green+src[j-dim].green+src[j+1-dim].green+src[j-1-dim].green)/6;
    dst[j].blue = (src[j].blue+src[j-1].blue+src[j+1].blue+src[j-dim].blue+src[j+1-dim].blue+src[j-1-dim].blue)/6;
}

for (j = dim+dim-1; j < dim*dim-1; j+=dim)
 {
    dst[j].red = (src[j].red+src[j-1].red+src[j-dim].red+src[j+dim].red+src[j-dim-1].red+src[j-1+dim].red)/6;
    dst[j].green = (src[j].green+src[j-1].green+src[j-dim].green+src[j+dim].green+src[j-dim-1].green+src[j-1+dim].green)/6;
    dst[j].blue = (src[j].blue+src[j-1].blue+src[j-dim].blue+src[j+dim].blue+src[j-dim-1].blue+src[j-1+dim].blue)/6;
}

i = i - (dim - 1);
for (j = dim; j < i; j+=dim)
 {
    dst[j].red = (src[j].red+src[j-dim].red+src[j+1].red+src[j+dim].red+src[j+1+dim].red+src[j-dim+1].red)/6;
    dst[j].green = (src[j].green+src[j-dim].green+src[j+1].green+src[j+dim].green+src[j+1+dim].green+src[j-dim+1].green)/6;
    dst[j].blue = (src[j].blue+src[j-dim].blue+src[j+1].blue+src[j+dim].blue+src[j+1+dim].blue+src[j-dim+1].blue)/6;
}

myJ = dim;

for (i = 1; i < dim-1; i++)
{
    for (j = 1; j < dim-1; j++)
    {
        myJ ++;
        dst[myJ].red = (src[myJ-1].red+src[myJ].red+src[myJ+1].red+src[myJ-dim-1].red+src[myJ-dim].red+src[myJ-dim+1].red+src[myJ+dim-1].red+src[myJ+dim].red+src[myJ+dim+1].red)/9;
        dst[myJ].green = (src[myJ-1].green+src[myJ].green+src[myJ+1].green+src[myJ-dim-1].green+src[myJ-dim].green+src[myJ-dim+1].green+src[myJ+dim-1].green+src[myJ+dim].green+src[myJ+dim+1].green)/9;
        dst[myJ].blue = (src[myJ-1].blue+src[myJ].blue+src[myJ+1].blue+src[myJ-dim-1].blue+src[myJ-dim].blue+src[myJ-dim+1].blue+src[myJ+dim-1].blue+src[myJ+dim].blue+src[myJ+dim+1].blue)/9;
    }
    myJ += 2;
}

有人可以解释这种优化是如何运作的吗？

Answer 1

由于您没有显示许多内容的定义（accumulate_sum，pixel_sum，RIDX等），因此很难确定。似乎代码的第一个版本按行和列迭代数据，第二个版本首先处理角点，然后处理边缘，然后处理其余数据。您根据周围的像素处理像素。角和边缘具有较少的相邻像素，因此处理它们更容易一些。打破这些特殊情况可以简化它们的代码。第二个版本然后unrolls处理其余数据时最里面的两个循环，这是唯一可能的，因为你已经消除了所有特殊情况（边和角）。

此更改是否实际上是＆＃34;优化＆＃34;留给读者的练习。您需要在两个版本上运行性能测试才能确定。即使第二个版本的效率更高，第一个版本的可读性也更高。

c代码优化平滑函数

1 个答案: