OpenCV convert to smaller data type: Max, or truncate/overflow?

Question

I need to pass a video buffer through a bandpass filter for a project I'm working on. The bandpass filter involves both an FFT and an IFFT, as well as a flattening of the frequencies outside of the pass band. In order to accomplish this, I need to have more room to store values than is available in the standard uchar type, so I am temporarily converting my video frames to CV_64F, with the plan to convert them back to CV_8UC3. Here is my question:

What is the behavior of OpenCV when converting cv::Mat data from a large data type to a smaller one? More specifically, if I have values in a cv::Mat object with CV_64F data that are larger than the maximum possible value for a single data point in CV_8U, does OpenCV set the target data to the maximum possible value for that data type, or do we have an overflow or truncating problem?

berak · Accepted Answer · 2014-07-18 06:11:49Z

2

Just wrote a simple program to figure out how this works:

#include <iostream>
#include <opencv2/imgproc/imgproc.hpp>
using namespace std;
int main() {
    cv::Mat img1(cv::Size(2,2), CV_64F);
    img1 = cv::Scalar(5000000);
    cv::Mat img2;
    img1.convertTo(img2, CV_8U);
    for(int i = 0; i < img2.rows; i++) {
        for(int j = 0; j < img2.cols; j++) {
            unsigned char tmp = img2.at<uchar>(i, j);
            cout << (int)tmp << endl;
        }
    }
    return 0;
}

Compiled it with pkg-config --cflags --libs opencv and ran it. The output was 255 on 4 lines.

Conclusion: OpenCV, when converting from larger data types to smaller ones, will set the smaller data type value to the max possible if the source data is a larger value. In other words, the value 500,000, which is larger than the max value of an unsigned char, will be set to 255 when converting to CV_8U.

edited Jul 18, 2014 at 6:11

berak

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answered Jul 18, 2014 at 3:34

Adam

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