OpenCV Display Pixels Value

Question

I have below code. When I run the program, unknown characters instead of pixel values comes to the screen. I want to display pixel values. How do I do this? Thank you.

#include <opencv2/opencv.hpp>
#include <opencv2/highgui.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <iostream>

using namespace cv;
using namespace std;

int main()
{
    Mat image = imread("/home/fd/baby.jpg");
    for( int i = 0 ; i < image.rows ; i++)
    {
        for( int j = 0 ; j < image.cols ; j++ )
        {
            if(image.type() == CV_8UC1)
            {
                image.at<uchar>(i,j) = 255;
            }
            else if(image.type() == CV_8UC3)
            {
                cout << image.at<Vec3b>(i,j)[0] << " " << image.at<Vec3b>(i,j)[1] << " " << image.at<Vec3b>(i,j)[2] << endl;

                image.at<Vec3b>(i,j)[0] = 255;
                image.at<Vec3b>(i,j)[1] = 255;
                image.at<Vec3b>(i,j)[2] = 255;

                cout << image.at<Vec3b>(i,j)[0] << " " << image.at<Vec3b>(i,j)[1] << " " << image.at<Vec3b>(i,j)[2] << endl;
            }
            else
            {
                cout << "Anknown image format" << endl;
                return 0;
            }
        }
    }
    imshow("Result İmage", image);
    waitKey(0);
}

This is the result screen:

enter image description here

possible duplicate of [Why is std::cout not printing the correct value for my int8\_t number?](http://stackoverflow.com/questions/7587782/why-is-stdcout-not-printing-the-correct-value-for-my-int8-t-number) — Alan Stokes, Apr 25 '15 at 13:24
But you know picture of pixels have values between 0 and 255. So I think my problem is not signed or unsigned pixel value. — fdas, Apr 25 '15 at 13:43
`cout << int(uchar_value);` but better even, just print the whole Mat: `cout << img << endl;` — berak, Apr 25 '15 at 13:56
It's not about signed or unsigned. Any flavour of `char` is printed as a single character not a number by default. Promoting to a bigger type prevents this. (If `cout << 'A'` printed 65 it would cause confusion. Blame backward compatibility.) — Alan Stokes, Apr 25 '15 at 14:23
Thank you @berak, I solved my problem wity your answer. Thank you. cout << int(image.at(i,j)[0]) — fdas, Apr 25 '15 at 15:02
possible duplicate of [Why "cout" works weird for "unsigned char"?](http://stackoverflow.com/questions/21374773/why-cout-works-weird-for-unsigned-char) — herohuyongtao, Apr 26 '15 at 02:06
@AlanStokes `Vec3b` is `Vec`, not `Vec`. So it's not a duplicate to what you link. :) — herohuyongtao, Apr 26 '15 at 02:08

kmdreko · Answer 1 · 2016-06-21T19:32:40.683

Cast each of the outputs to an integer

<< image.at<Vec3b>(i,j)[0] ...

change to

<< (int)image.at<Vec3b>(i,j)[0] ...

You are printing a char (or probably unsigned char) which is printed through the stream as a single character (which at 255 looks like what you see). Cast to int forces it to display the numerical representation of the value.

The other answers changing the image.at<type> changes how the raw data is interpreted; don't do that. They must be interpreted properly.

score 0 · Answer 2 · edited Nov 04 '15 at 10:26

0

Change the

image.at<Vec3b>(i, j)

to

image.at<int>(i, j)

or

image.at<double>(i, j)

or

image.at<float>(i, j)

to print the values instead of characters

edited Nov 04 '15 at 10:26

Willi Mentzel

27,862
20
113
121

answered Nov 04 '15 at 09:14

Santhosh Nagasanthosh

13
1
2

score 0 · Accepted Answer · answered Nov 08 '19 at 20:20

0

Vec3b px= image.at(x,y);

cout << "value: ("<<(int)px.val[0]<<", "<<(int)px.val[1]<<", "<<(int)px.val[2]<<")" << endl;

This works form me.

answered Nov 08 '19 at 20:20

user3476405

26
1
2

score -2 · Answer 4 · edited Jun 21 '16 at 19:18

-2

You are displaying characters try converting using function image.at<int>(j,i);

edited Jun 21 '16 at 19:18

Avijit Dasgupta

2,055
3
22
36

answered Oct 21 '15 at 05:34

Santhosh Nagasanthosh

13
1
2

OpenCV Display Pixels Value

4 Answers4