在opencv中查找正方形:访问冲突读取位置

时间:2013-09-21 08:15:31

标签: c++ opencv

我尝试在C ++示例中运行opencv方向的squares.cpp,一切正常,但是当程序达到这一点时:aboutPolyDP(Mat(contours [i]),approx,arcLength(Mat(contours [i]) ,true)* 0.02,true); 我得到例外说: FindRectangle.exe中0x61163C77(opencv_imgproc244d.dll)的未处理异常:0xC0000005:访问冲突读取位置0x030F9000。 我做任何事来解决这个问题,但我做不到。 我在visual studio 2012中使用32位处理运行它。请帮助!!!!!!!!!!

static double angle( Point pt1, Point pt2, Point pt0 )
{
double dx1 = pt1.x - pt0.x;
double dy1 = pt1.y - pt0.y;
double dx2 = pt2.x - pt0.x;
double dy2 = pt2.y - pt0.y;
return (dx1*dx2 + dy1*dy2)/sqrt((dx1*dx1 + dy1*dy1)*(dx2*dx2 + dy2*dy2) + 1e-10);
}

// returns sequence of squares detected on the image.
// the sequence is stored in the specified memory storage
static void findSquares( const Mat& image, vector<vector<Point> >& squares )
{
squares.clear();

Mat pyr, timg, gray0(image.size(), CV_8U), gray;

// down-scale and upscale the image to filter out the noise
pyrDown(image, pyr, Size(image.cols/2, image.rows/2));
pyrUp(pyr, timg, image.size());
vector<vector<Point> > contours;

// find squares in every color plane of the image
for( int c = 0; c < 3; c++ )
{
    int ch[] = {c, 0};
    mixChannels(&timg, 1, &gray0, 1, ch, 1);

    // try several threshold levels
    for( int l = 0; l < N; l++ )
    {
        // hack: use Canny instead of zero threshold level.
        // Canny helps to catch squares with gradient shading
        if( l == 0 )
        {
            // apply Canny. Take the upper threshold from slider
            // and set the lower to 0 (which forces edges merging)
            Canny(gray0, gray, 0, thresh, 5);
            // dilate canny output to remove potential
            // holes between edge segments
            dilate(gray, gray, Mat(), Point(-1,-1));
        }
        else
        {
            // apply threshold if l!=0:
            //     tgray(x,y) = gray(x,y) < (l+1)*255/N ? 255 : 0
            gray = gray0 >= (l+1)*255/N;
        }

        // find contours and store them all as a list
        findContours(gray, contours, CV_RETR_LIST, CV_CHAIN_APPROX_SIMPLE);

        vector<Point> approx ;

        // test each contour
        for( size_t i = 0; i < contours.size(); i++ )
        {
            // approximate contour with accuracy proportional
            // to the contour perimeter
            approxPolyDP(Mat(contours[i]), approx, arcLength(Mat(contours[i]), true)*0.02, true);

            // square contours should have 4 vertices after approximation
            // relatively large area (to filter out noisy contours)
            // and be convex.
            // Note: absolute value of an area is used because
            // area may be positive or negative - in accordance with the
            // contour orientation
            if( approx.size() == 4 &&
                fabs(contourArea(Mat(approx))) > 1000 &&
                isContourConvex(Mat(approx)) )
            {
                double maxCosine = 0;

                for( int j = 2; j < 5; j++ )
                {
                    // find the maximum cosine of the angle between joint edges
                    double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
                    maxCosine = MAX(maxCosine, cosine);
                }

                // if cosines of all angles are small
                // (all angles are ~90 degree) then write quandrange
                // vertices to resultant sequence
                if( maxCosine < 0.3 )
                    squares.push_back(approx);
            }
            else{
                approx.clear();
            }
        }
    }
}


// the function draws all the squares in the image
 static void drawSquares( Mat& image, const vector<vector<Point> >& squares )
{
for( size_t i = 0; i < squares.size(); i++ )
{
    const Point* p = &squares[i][0];
    int n = (int)squares[i].size();
    polylines(image, &p, &n, 1, true, Scalar(0,255,0), 3, CV_AA);
}

imshow(wndname, image);
}

0 个答案:

没有答案