Understanding the copy done by memcpy()

Question

I have to create an image which has two times the number of columns to that of the original image. Therefore, I have kept the width of the new image two times to that of the original image.

Although this is a very simple task and I have already done it but I am wondering about the strange results obtained by doing this task using memcpy().

My code:

int main()
{

    Mat image = imread("pikachu.png", 1);
    int columns = image.cols;
    int rows = image.rows;

    Mat twoTimesImage(image.rows, 2 * image.cols, CV_8UC3, Scalar(0));

    unsigned char *pSingleImg = image.data;
    unsigned char *ptwoTimes = twoTimesImage.data;

    size_t memsize = 3 * image.rows*image.cols;

    memcpy(ptwoTimes , pSingleImg, memsize);
    memcpy(ptwoTimes + memsize, pSingleImg, memsize);

    cv::imshow("two_times_image.jpg", twoTimesImage);

    return 0;
}

Original Image:

Result

Expected Results:

Question: When the resulting image is just two times to that of the original image then, how come 4 original images are getting copied to the new image? Secondly, the memcpy() copies the continous memory location in a row-wise fashion so, according to that I should get an image which is shown in the "Expected results".

what is this line `size_t memsize = 3 * image.rows*image.cols;`? specifically, `3`? — Sourav Ghosh, Jul 28 '15 at 10:16
@SouravGhosh I suspect it's 3 bytes per pixel for the R, G, B components. — The Paramagnetic Croissant, Jul 28 '15 at 10:17
@TheParamagneticCroissant in that case, isn't like `twoTimesImage` is missing a multiplier 2 for `image.rows`. Pure guess, though. — Sourav Ghosh, Jul 28 '15 at 10:19
It is a color image and hence contain R,G and B channel. Therefore, each pixel consists for 3 bytes. — skm, Jul 28 '15 at 10:19
@SouravGhosh: As, I mentioned that I have increased the width only. If you multiply the `image.rows` by 2, you will increase the height. I have tried that too but the results will remain weird. So, to keep the question simple, I have only increase the width. — skm, Jul 28 '15 at 10:21
Are you sure that the data matrix layout of the memory is contiguous? — LPs, Jul 28 '15 at 12:44
@Lps: Yes, I have already provided an answer below which suggests the same. — skm, Jul 28 '15 at 13:06

score 9 · Accepted Answer · answered Jul 28 '15 at 10:41

9

The left cat consists of the odd numbered lines and the right cat consists of the even numbered lines of the original picture. This is then doubled, so that there are two more cats underneath. The new cats have half the number of lines of the original cat.

The new picture is laid out like this:

line 1  line 2
line 3  line 4
line 5  line 6
...     
line n-1 line n
line 1  line 2
line 3  line 4
line 5  line 6
...     
line n-1 line n

answered Jul 28 '15 at 10:41

Klas Lindbäck

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12

The "cat" is clearly a [pikachu](https://duckduckgo.com/?q=pikachu&t=opera&ia=images). – nwp Jul 28 '15 at 10:50

score 5 · Answer 2 · answered Jul 28 '15 at 11:07

the answer provided by "Klas Lindbäck" is absolutely correct. Just to provide more clarity to someone who might have a similar confusion, I am writing this answer. I created an image with odd rows consisting of red color and even rows consisting of blue color.

Then, I used the code given in my original post. As, expected by the answer of "Klas Lindbäck", the red color came into the first coloumn and the blue color came into the second column.

Original image:

Resulting image:

Understanding the copy done by memcpy()

2 Answers2

Linked