2013-12-11 03:42:00 -0600 | received badge | ● Necromancer (source) |
2013-12-11 02:24:43 -0600 | answered a question | Gpu version of filter2D Hi, The statement that you are executing has a syntax error because the arguments do not match the defination of the function: cv::gpu::filter2D(gpuPatch, gpuResponse, CV_32F, m_model.weights[i], cv::Point(-1, -1), 0, cv::BORDER_CONSTANT); Try executing the following statement, you wont get a compile time error: cv::gpu::filter2D(gpuPatch, gpuResponse, CV_32F, m_model.weights[i], cv::Point(-1, -1),cv::BORDER_CONSTANT); |
2013-11-27 21:48:47 -0600 | received badge | ● Editor (source) |
2013-11-26 00:30:11 -0600 | asked a question | Facing output difference while using different OpenCv Versions Previously, I was using OpenCV 2.2 and now I am using OpenCV 2.4.5. The output of the cvCvtColor() function is different when converting from RGB to HSV, and the difference is in H(Hue). The input of RGB is (211,216,255) and the output is: •Using OpenCV 2.2: (116,44,255) •Using OpenCV 2.4.5: (117,44,255) The output difference is found in the Hue array (116 | 117). When I calculated it manually using the formula mentioned in the documentation, the output is 116.590909. So the assumption here is that OpenCV 2.4.5 rounds off the value and OpenCV 2.2 discards the values after the decimal point. Can anyone please let me know why this difference happens? |