I m new with C++, andtries figuring the what this line means:
cur_rect = cv::Rect(cur_rect) & cv::Rect(0, 0, mat->cols, mat->rows); // here
if( cv::Rect(cur_rect) == cv::Rect() ) //here
{
.......
}
I m new with C++, andtries figuring the what this line means:
cur_rect = cv::Rect(cur_rect) & cv::Rect(0, 0, mat->cols, mat->rows); // here
if( cv::Rect(cur_rect) == cv::Rect() ) //here
{
.......
}
The Rect & Rect
part intersection two rectangles and given a non-empty rectangle Back when the two submissions coordinated.
因此,你可以将结果与<代码>Rect()相比较,看看是否存在交叉点。 页: 1
Sources:
http://opencv.willowgarage.com/documentation/cpp/core_basic_ructs.html?highlight=rect
替代执行,比照更清洁:
// crop cur_rect to rectangle with matrix mat size:
cur_rect &= cv::Rect(0, 0, mat->cols, mat->rows);
if (cur_rect.area() == 0) {
// result is empty
...
}
我假定,<代码>cv:Rect(......)(或其家属)方法回归了一个试金字。 您不理解的界线,我假定是超负荷运营商(=
),对中继器进行比较。
但是,我在此作许多假设,因为我没有<代码>cv类别的代码。
关于<代码>&超载运营商——假设这是在交叉或结合中进行的。 如果没有守则,就很难说。
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