填充后存储图像的问题

如何解决填充后存储图像的问题

我正在使用5x5窗口执行简单的遮罩操作。遵循本文Problem with Operation on Border pixels of an image（第二种解决方案）中公认的解决方案，我创建了一个图像img_clamp，该图像可以用作具有4个额外行和4个额外列的缓冲区。请在下面找到示例代码。

int main(int argc,char** argv)
{
    Mat input = imread("C:/Users/20181217/Desktop/images/imgs/den_check.png");
    //input.rows = 256,input.cols =512
    Mat output = (input.rows,input.cols,input.type()); //row and col = 256 and 512 
    
        
    //number of additional rows and columns you woluld ike on each side
    int top,left,right,bottom;
    top = 2;
    left = 2;
    right = 2;
    bottom = 2;
    
    //define new image with additional borders
    Mat img_clamp(input.rows + 4,input.cols + 4,CV_8UC3);  
        
    
    //if you want to replicate the border of the image
    copyMakeBorder(input,img_clamp,top,bottom,BORDER_REPLICATE);
    //img_clamp row and col size : 260 and 516


//Now you can access the image without having to worry about the borders as shown below

//start iterationg from the 2nd row till 258th row (this leaves 1,2,259 and 260th rows for the out of bounds access by the 5x5 window) 
for(int i=2;i<input.rows-2;i++)
 {
    //start iteration from the 2nd col till 514th col (this leaves 1,515 and 516th cols for the out of bounds access by the 5x5 window)
  for(int j=2;i<input.cols-2;i++)
   {
     
    temp_red = img_clamp.at<Vec3b>[0](i-2,j-2) + img_clamp.at<Vec3b>[0](i-2,j+2) + img_clamp.at<Vec3b>[0](i+2,j-2) + img_clamp.at<Vec3b>[0](i+2,j+2);
    temp_green = img_clamp.at<Vec3b>[1](i-2,j-2) + img_clamp.at<Vec3b>[1](i-2,j+2) + img_clamp.at<Vec3b>[1](i+2,j-2) + img_clamp.at<Vec3b>[1](i+2,j+2);
    temp_blue = img_clamp.at<Vec3b>[2](i-2,j-2) + img_clamp.at<Vec3b>[2](i-2,j+2) + img_clamp.at<Vec3b>[2](i+2,j-2) + img_clamp.at<Vec3b>[2](i+2,j+2);
     
     ...
     
     //store values in the output image which has the same the size as input image (i.e 256 and 512 (rows and cols))
     output.at<Vec3b>[0](i-2,j-2) =temp_red
     output.at<Vec3b>[1](i-2,j-2) =temp_green
     output.at<Vec3b>[2](i-2,j-2) =temp_blue
         
         
    }
  }

//Code for checking the output is matching with the golden data(ideal image)<---- the ideal image is from a different language halide,My goal is to replicate the same logic in c++

Mat diff = abs(ideal-output);

    Mat diff = abs(small_img - ideal);
    //cout << diff;
    int r,g,b,t,n,r_b,g_b,b_b;
    r = 0; //faulty red
    g = 0;//faulty green
    b = 0;//faulty blue
    n = 393216; //total number of pixels
    //_b  indicates the border pixels
    r_b = 0;
    b_b = 0;
    g_b = 0;

    t = 1;//threshold for the difference

          // printing out the difference between final image and ideal image (if any)
    for (int i = 0; i < input.cols; i++)//512 col size x
    {
        for (int j = 0; j < input.rows; j++)//256 row size y
        {
            if ((int)diff.at<Vec3b>(j,i)[0] > t)
            {
                if (i == 508 || i == 509 || i == 510 || i == 511 || j == 252 || j == 253 || j == 254 || j == 255) //border pixels(right most and bottom most of the image)
                    r_b++;//increment if its one of the border pixels
                //printing the pixel position which has the wrong value
                cout << "problem at (" << j << "," << i << ")  of red :" << (int)output.at<Vec3b>(j,i)[0] << ",expected value:" << (int)ideal.at<Vec3b>(j,i)[0] << endl;
                r++;//increment if the red pixel is faulty and not matching with the ideal image
                n--;
            }
            if ((int)diff.at<Vec3b>(j,i)[1] > t)
            {
                if (i == 508 || i == 509 || i == 510 || i == 511 || j == 252 || j == 253 || j == 254 || j == 255)
                    g_b++;
                
                g++;//increment if the green pixel is faulty and not matching with the ideal image
                n--;

            }
            if ((int)diff.at<Vec3b>(j,i)[2] > t)
            {
                if (i == 508 || i == 509 || i == 510 || i == 511 || j == 252 || j == 253 || j == 254 || j == 255)
                    b_b++;              

                b++;//increment if the blue pixel is faulty and not matching with the ideal image
                n--;
            }

        }
    }
    cout << endl << endl << "for a threshold of difference greater than " << t << endl << "total faulty pixels are(R,G,B) :(" << r << " " << g << " " << b << ")"<<endl;
    cout << " The border pixels present in the faulty pixels are :(" << r_b << " " << g_b << " " << b_b << endl << endl;
    cout<<" correct pixels that match with the ideal image = " << n;
    
    return 0;
}

我想将输出存储为常规图像**（未填充）**。 除了最后四行和四列之外，一切都按照逻辑进行。控制台的输出为

我知道它在存储时在(i-2,j-2)上出了问题，但是我找不到解决方法。

任何有关如何解决它的帮助/建议都将受到赞赏。

预先感谢

解决方法

问题出在for loop的边界 代替这个

for(int i=2;i<input.rows-2;i++)//size =256
 {
    //start iteration from the 2nd col till 514th col (this leaves 1,2,515 and 516th cols for the out of bounds access by the 5x5 window)
  for(int j=2;i<input.cols-2;i++)//size =512

应该是

for(int i=2;i<img_clamp.rows-2;i++)//size = 260
 {
    //start iteration from the 2nd col till 514th col (this leaves 1,515 and 516th cols for the out of bounds access by the 5x5 window)
  for(int j=2;i<img_clamp.cols-2;i++)//size = 516