C++ OpenCV:未能正确使用LineIterator，给出了无意义的值

我在下面的程序中遇到了一个非常奇怪的行为。我基本上是在一个图像上迭代2个点，然后收集可变行中的所有像素值

void IterateLine( const Mat& image, vector<ushort>& linePixels, Point p2, Point p1, int *count1 )
{
     //Am I getting correct image? 
    cout<<"Pixel values in image in IterateLine"<<endl; 

    cout<<Mat(image)<<endl; //Yes this shows meaningful values indeed 


 //hard coding pixel coordinates for debugging. Prints meaningful pixel values at end  
p1.x=32;
p1.y=66;

p2.x=230;
p2.y=86;

//but If I do not use hard coded values as above, then it always prints 114 



//Am I getting correct pixel coordinates? Let us print   
cout<<"First coordinates received in IterateLine()  are given below"<<endl;
cout<<p1.x<<'\t'<<p1.y<<endl; 

cout<<"Second coordinates received in IterateLine()  are given below"<<endl;
cout<<p2.y<<'\t'<<p2.y<<endl;

//The above indeed prints meaningful coordinates, but only when pixel values are hard coded   

    LineIterator it(image, p2,p1, 8);

    for(int i = 0; i < it.count; i++, it++){
        linePixels.push_back(image.at<ushort>(it.pos()));
    }

   *count1=it.count; 


    cout<<"No. of pixels in IterateLLine"<<endl;

    cout<<count1[0]<<endl; 


cout<<"pixel values in  IterateLine are"<<endl;

    for(int i=0;i<count1[0];i++)

    {

        cout<<linePixels[i]<<endl;
    }



}

我正确地传递了main中的两个点，因为我在传递到IterateLine和IterateLine内部之前进行了验证。我看到两者是平等的

void IterateLine( const Mat& image, vector<ushort>& linePixels, Point p2, Point p1, int *count1 )
{
     //Am I getting correct image? 
    cout<<"Pixel values in image in IterateLine"<<endl; 

    cout<<Mat(image)<<endl; //Yes this shows meaningful values indeed 


 //hard coding pixel coordinates for debugging. Prints meaningful pixel values at end  
p1.x=32;
p1.y=66;

p2.x=230;
p2.y=86;

//but If I do not use hard coded values as above, then it always prints 114 



//Am I getting correct pixel coordinates? Let us print   
cout<<"First coordinates received in IterateLine()  are given below"<<endl;
cout<<p1.x<<'\t'<<p1.y<<endl; 

cout<<"Second coordinates received in IterateLine()  are given below"<<endl;
cout<<p2.y<<'\t'<<p2.y<<endl;

//The above indeed prints meaningful coordinates, but only when pixel values are hard coded   

    LineIterator it(image, p2,p1, 8);

    for(int i = 0; i < it.count; i++, it++){
        linePixels.push_back(image.at<ushort>(it.pos()));
    }

   *count1=it.count; 


    cout<<"No. of pixels in IterateLLine"<<endl;

    cout<<count1[0]<<endl; 


cout<<"pixel values in  IterateLine are"<<endl;

    for(int i=0;i<count1[0];i++)

    {

        cout<<linePixels[i]<<endl;
    }



}

但我得到的线像素总是114

void IterateLine( const Mat& image, vector<ushort>& linePixels, Point p2, Point p1, int *count1 )
{
     //Am I getting correct image? 
    cout<<"Pixel values in image in IterateLine"<<endl; 

    cout<<Mat(image)<<endl; //Yes this shows meaningful values indeed 


 //hard coding pixel coordinates for debugging. Prints meaningful pixel values at end  
p1.x=32;
p1.y=66;

p2.x=230;
p2.y=86;

//but If I do not use hard coded values as above, then it always prints 114 



//Am I getting correct pixel coordinates? Let us print   
cout<<"First coordinates received in IterateLine()  are given below"<<endl;
cout<<p1.x<<'\t'<<p1.y<<endl; 

cout<<"Second coordinates received in IterateLine()  are given below"<<endl;
cout<<p2.y<<'\t'<<p2.y<<endl;

//The above indeed prints meaningful coordinates, but only when pixel values are hard coded   

    LineIterator it(image, p2,p1, 8);

    for(int i = 0; i < it.count; i++, it++){
        linePixels.push_back(image.at<ushort>(it.pos()));
    }

   *count1=it.count; 


    cout<<"No. of pixels in IterateLLine"<<endl;

    cout<<count1[0]<<endl; 


cout<<"pixel values in  IterateLine are"<<endl;

    for(int i=0;i<count1[0];i++)

    {

        cout<<linePixels[i]<<endl;
    }



}

但是如果我在IterateLine中硬编码点的开始和结束，那么我会得到有意义的值

void IterateLine( const Mat& image, vector<ushort>& linePixels, Point p2, Point p1, int *count1 )
{
     //Am I getting correct image? 
    cout<<"Pixel values in image in IterateLine"<<endl; 

    cout<<Mat(image)<<endl; //Yes this shows meaningful values indeed 


 //hard coding pixel coordinates for debugging. Prints meaningful pixel values at end  
p1.x=32;
p1.y=66;

p2.x=230;
p2.y=86;

//but If I do not use hard coded values as above, then it always prints 114 



//Am I getting correct pixel coordinates? Let us print   
cout<<"First coordinates received in IterateLine()  are given below"<<endl;
cout<<p1.x<<'\t'<<p1.y<<endl; 

cout<<"Second coordinates received in IterateLine()  are given below"<<endl;
cout<<p2.y<<'\t'<<p2.y<<endl;

//The above indeed prints meaningful coordinates, but only when pixel values are hard coded   

    LineIterator it(image, p2,p1, 8);

    for(int i = 0; i < it.count; i++, it++){
        linePixels.push_back(image.at<ushort>(it.pos()));
    }

   *count1=it.count; 


    cout<<"No. of pixels in IterateLLine"<<endl;

    cout<<count1[0]<<endl; 


cout<<"pixel values in  IterateLine are"<<endl;

    for(int i=0;i<count1[0];i++)

    {

        cout<<linePixels[i]<<endl;
    }



}

我不明白这里出了什么问题？有些东西我不知道

void IterateLine( const Mat& image, vector<ushort>& linePixels, Point p2, Point p1, int *count1 )
{
     //Am I getting correct image? 
    cout<<"Pixel values in image in IterateLine"<<endl; 

    cout<<Mat(image)<<endl; //Yes this shows meaningful values indeed 


 //hard coding pixel coordinates for debugging. Prints meaningful pixel values at end  
p1.x=32;
p1.y=66;

p2.x=230;
p2.y=86;

//but If I do not use hard coded values as above, then it always prints 114 



//Am I getting correct pixel coordinates? Let us print   
cout<<"First coordinates received in IterateLine()  are given below"<<endl;
cout<<p1.x<<'\t'<<p1.y<<endl; 

cout<<"Second coordinates received in IterateLine()  are given below"<<endl;
cout<<p2.y<<'\t'<<p2.y<<endl;

//The above indeed prints meaningful coordinates, but only when pixel values are hard coded   

    LineIterator it(image, p2,p1, 8);

    for(int i = 0; i < it.count; i++, it++){
        linePixels.push_back(image.at<ushort>(it.pos()));
    }

   *count1=it.count; 


    cout<<"No. of pixels in IterateLLine"<<endl;

    cout<<count1[0]<<endl; 


cout<<"pixel values in  IterateLine are"<<endl;

    for(int i=0;i<count1[0];i++)

    {

        cout<<linePixels[i]<<endl;
    }



}

请帮忙

void IterateLine( const Mat& image, vector<ushort>& linePixels, Point p2, Point p1, int *count1 )
{
     //Am I getting correct image? 
    cout<<"Pixel values in image in IterateLine"<<endl; 

    cout<<Mat(image)<<endl; //Yes this shows meaningful values indeed 


 //hard coding pixel coordinates for debugging. Prints meaningful pixel values at end  
p1.x=32;
p1.y=66;

p2.x=230;
p2.y=86;

//but If I do not use hard coded values as above, then it always prints 114 



//Am I getting correct pixel coordinates? Let us print   
cout<<"First coordinates received in IterateLine()  are given below"<<endl;
cout<<p1.x<<'\t'<<p1.y<<endl; 

cout<<"Second coordinates received in IterateLine()  are given below"<<endl;
cout<<p2.y<<'\t'<<p2.y<<endl;

//The above indeed prints meaningful coordinates, but only when pixel values are hard coded   

    LineIterator it(image, p2,p1, 8);

    for(int i = 0; i < it.count; i++, it++){
        linePixels.push_back(image.at<ushort>(it.pos()));
    }

   *count1=it.count; 


    cout<<"No. of pixels in IterateLLine"<<endl;

    cout<<count1[0]<<endl; 


cout<<"pixel values in  IterateLine are"<<endl;

    for(int i=0;i<count1[0];i++)

    {

        cout<<linePixels[i]<<endl;
    }



}

像素坐标未硬编码时输出

void IterateLine( const Mat& image, vector<ushort>& linePixels, Point p2, Point p1, int *count1 )
{
     //Am I getting correct image? 
    cout<<"Pixel values in image in IterateLine"<<endl; 

    cout<<Mat(image)<<endl; //Yes this shows meaningful values indeed 


 //hard coding pixel coordinates for debugging. Prints meaningful pixel values at end  
p1.x=32;
p1.y=66;

p2.x=230;
p2.y=86;

//but If I do not use hard coded values as above, then it always prints 114 



//Am I getting correct pixel coordinates? Let us print   
cout<<"First coordinates received in IterateLine()  are given below"<<endl;
cout<<p1.x<<'\t'<<p1.y<<endl; 

cout<<"Second coordinates received in IterateLine()  are given below"<<endl;
cout<<p2.y<<'\t'<<p2.y<<endl;

//The above indeed prints meaningful coordinates, but only when pixel values are hard coded   

    LineIterator it(image, p2,p1, 8);

    for(int i = 0; i < it.count; i++, it++){
        linePixels.push_back(image.at<ushort>(it.pos()));
    }

   *count1=it.count; 


    cout<<"No. of pixels in IterateLLine"<<endl;

    cout<<count1[0]<<endl; 


cout<<"pixel values in  IterateLine are"<<endl;

    for(int i=0;i<count1[0];i++)

    {

        cout<<linePixels[i]<<endl;
    }



}

我的主要工作如下：

void IterateLine( const Mat& image, vector<ushort>& linePixels, Point p2, Point p1, int *count1 )
{
     //Am I getting correct image? 
    cout<<"Pixel values in image in IterateLine"<<endl; 

    cout<<Mat(image)<<endl; //Yes this shows meaningful values indeed 


 //hard coding pixel coordinates for debugging. Prints meaningful pixel values at end  
p1.x=32;
p1.y=66;

p2.x=230;
p2.y=86;

//but If I do not use hard coded values as above, then it always prints 114 



//Am I getting correct pixel coordinates? Let us print   
cout<<"First coordinates received in IterateLine()  are given below"<<endl;
cout<<p1.x<<'\t'<<p1.y<<endl; 

cout<<"Second coordinates received in IterateLine()  are given below"<<endl;
cout<<p2.y<<'\t'<<p2.y<<endl;

//The above indeed prints meaningful coordinates, but only when pixel values are hard coded   

    LineIterator it(image, p2,p1, 8);

    for(int i = 0; i < it.count; i++, it++){
        linePixels.push_back(image.at<ushort>(it.pos()));
    }

   *count1=it.count; 


    cout<<"No. of pixels in IterateLLine"<<endl;

    cout<<count1[0]<<endl; 


cout<<"pixel values in  IterateLine are"<<endl;

    for(int i=0;i<count1[0];i++)

    {

        cout<<linePixels[i]<<endl;
    }



}

int main()

{

     Mat img = imread("sir.png", CV_LOAD_IMAGE_ANYCOLOR | CV_LOAD_IMAGE_ANYDEPTH);


 vector<Point> points1;

 vector<ushort>linePixels;

 //Draw the line, and collect end and final point coordinates in points1 

      drawline(img, points1); 

//now check the collected points

cout<<"First coordinates passed to IterateLine() in main() are given below"<<endl;
cout<<points1[0].x<<'\t'<<points1[0].y<<endl; 


cout<<"Second coordinates passed to IterateLine() in main() are given below"<<endl;
cout<<points1[1].x<<'\t'<<points1[1].y<<endl; 


IterateLine( img, linePixels, points1[1], points1[0], t );   

return 0 ; 
}  

---

但这不是很有效的方法，所以我仍然在寻找最好的方法

@berak你的意思是我正在获取我绘制的“线”的像素值？@berak那么我如何使这条线透明，以便IterateLine看不到它？只有人类才能看到这一点line@berak请查看我的更新，我已经显示了我的drawline@berak我单击的两个点存储在vector&points中，然后我在迭代线中使用它；你可以直接复制图像，而不是读取图像两次。它将比读取磁盘上的数据更有效。不过，要注意复制Mat的不同功能。有些只复制标题，仍然引用相同的基础数据，有些则真正复制标题和新Mat对象中的数据。

void IterateLine( const Mat& image, vector<ushort>& linePixels, Point p2, Point p1, int *count1 )
{
     //Am I getting correct image? 
    cout<<"Pixel values in image in IterateLine"<<endl; 

    cout<<Mat(image)<<endl; //Yes this shows meaningful values indeed 


 //hard coding pixel coordinates for debugging. Prints meaningful pixel values at end  
p1.x=32;
p1.y=66;

p2.x=230;
p2.y=86;

//but If I do not use hard coded values as above, then it always prints 114 



//Am I getting correct pixel coordinates? Let us print   
cout<<"First coordinates received in IterateLine()  are given below"<<endl;
cout<<p1.x<<'\t'<<p1.y<<endl; 

cout<<"Second coordinates received in IterateLine()  are given below"<<endl;
cout<<p2.y<<'\t'<<p2.y<<endl;

//The above indeed prints meaningful coordinates, but only when pixel values are hard coded   

    LineIterator it(image, p2,p1, 8);

    for(int i = 0; i < it.count; i++, it++){
        linePixels.push_back(image.at<ushort>(it.pos()));
    }

   *count1=it.count; 


    cout<<"No. of pixels in IterateLLine"<<endl;

    cout<<count1[0]<<endl; 


cout<<"pixel values in  IterateLine are"<<endl;

    for(int i=0;i<count1[0];i++)

    {

        cout<<linePixels[i]<<endl;
    }



}