Accessing each pixel of a Mat .at vs pointer arithmetic gives differnet result

c++

asked 2016-05-13 12:42:57 -0600

updated 2016-05-13 12:52:45 -0600

Today I stumbled across this (for me) weird thing, namely when I run this code

m.convertTo(tmp_mat, CV_32FC3, 1./255);
cvtColor(tmp_mat, tmp_mat, CV_RGB2XYZ, 3);
split(tmp_mat, xyz_channels);

y_channel = xyz_channels[1].clone();

int i,j;
uint8_t* p = y_channel.data;
for( i = 0; i < y_channel.rows; ++i)
{
    for ( j = 0; j < y_channel.cols; ++j)
    {
       std::cout << y_channel.at<float>(i,j) << std::endl;
       std::cout << saturate_cast<float>(p[i*y_channel.cols + j]) << std::endl;
    }
}

this is the output I get

So now I am confused - shouldn't it be same?

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Comments

CV_32FC3 implies mat.at<float>(r,c) , not mat.at<double>(r,c) (which would require CV_64F3C)

berak ( 2016-05-13 12:49:37 -0600 )edit

@berak Thanks for the hint but problem is same... Please check edited question.

carobnodrvo ( 2016-05-13 12:53:08 -0600 )edit

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answered 2016-05-13 22:05:07 -0600

tuannhtn

1500 ●10 ●19 http://4fire.wordpress...

updated 2016-05-14 03:11:17 -0600

You have to change the line:

 uint8_t* p = y_channel.data;

to:

float * p = (float*)y_channel.data;

and no need to use saturate_cast<>, just do like this:

std::cout << p[i*y_channel.cols + j] << std::endl;

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answered 2016-05-13 16:09:10 -0600

Tetragramm

7376 ●13 ●37

You have p as type uint8_t, when the data inside is of type float. You should declare it as a pointer of type float and cast the pointer you get from .data

As it is, you are accessing the 8 bits as a uint8, then converting that to a float, which is of course, wrong. You need to access the full 32 bits as a float.

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Accessing each pixel of a Mat .at vs pointer arithmetic gives differnet result

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Accessing each pixel of a Mat .at vs pointer arithmetic gives differnet result