Question

cv::Mat x(2,2,CV_32F);
x.at<float>(0,0)=0.7;
x.at<float>(0,1) = 0.8;
x.at<float>(1,0) = 0.72;
x.at<float>(1,1) = 0.68;
x.convertTo( x, CV_8U, 255, 0  );
std::cout << x.at<int>(0,0)  << std::endl;
std::cout << x.at<int>(0,1)  << std::endl;
std::cout << x.at<float>(1,0)  << std::endl; // I deliberately put <float> just to see what happens
std::cout << x.at<int>(1,1)  << std::endl;

输出结果为：

-1380397902 1 1.54154e-40 0

我期待输出如下：

178 204 184 173

我做错了什么？

Answer 1

如果您的Mat现在是CV_8U，则必须以

的形式访问它

x.at<uchar>(1,1)

下一个问题是正确打印出数字，cout，看到一个字母，会尝试打印一个字母，所以你必须把它转换为int：

std::cout << int(x.at<uchar>(0,0))  << std::endl;

然后，如果你的垫子是2x2，你不能访问2,2的元素，这是超出范围，并将导致UB。！

（在c ++中，我们从0..n-1索引，对吗？）

所以，这里是更正后的例子：

cv::Mat x(2,2,CV_32F);
x.at<float>(0,0)=0.7;
x.at<float>(1,0) = 0.8;
x.at<float>(0,1) = 0.72;
x.at<float>(1,1) = 0.68;
Mat y;

x.convertTo( y, CV_8U, 255, 0  );
cout << int(y.at<uchar>(0,0))  << std::endl;
cout << int(y.at<uchar>(1,0))  << std::endl;
cout << int(y.at<uchar>(0,1))  << std::endl;
cout << int(y.at<uchar>(1,1))  << std::endl;

cerr << x << endl;
cerr << y << endl;

178
204
184
173
[0.69999999, 0.72000003;
  0.80000001, 0.68000001]
[178, 184;
  204, 173]

OpenCV C ++：使用converTo从CV_32F转换到CV8U会产生意外的值

1 个答案: