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能够理解的是,按照C++代码线,
原标题:can t understand following c++ code line
  • 时间:2012-05-10 09:45:02
  •  标签:
  • c++
  • opencv

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
{
.......
}
最佳回答

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

如何容易发现开放式Cv中的2个代谢 inter?

Edit

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
    ...
}
问题回答

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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