Unable to perform assignment because the size of the left side is 128-by-384 and the size of the right side is 128-by-128.

oriImage = imread('lena.bmp');

Watermark = imread('new_logo128.bmp');

[M,N]=size(oriImage);

[M1,N1] = size(Watermark);

watermarkImage = zeros(M,N);

for i=1:M/M1

for j = 1:N/N1

watermarkImage((i-1)*M1+1:(i-1)*M1+M1,(j-1)*N1+1:(j-1)*N1+N1)=Watermark(:,:,1);

end

end

figure

subplot(2,3,1);imshow(oriImage);title("original Image")

subplot(2,3,2);imshow(watermarkImage);title("watermark Image")

bitplane=1;

for i=1:M

for j = 1:N

watermarkImage(i,j) = bitset(oriImage(i,j),bitplane,watermarkImage(i,j));

end

end

subplot(2,3,3);imshow(watermarkImage*255);title('watermark image')

[Tony]

On which line of code did this error occur?

[frank]

Error in watermark (line 9)

watermarkImage((i-1)*M1+1:(i-1)*M1+M1,(j-1)*N1+1:(j-1)*N1+N1)=Watermark(:,:,1);

Answer 1

[DGM]

I'm guessing here, since I don't know what your images are, but it appears that they are both RGB images. In either case, you're misusing size(). If you request K scalar output arguments from size(), the last output is not the size of the Kth dimension. It's the product of the sizes of the Kth dimension and any trailing dimensions. Unless you have a need for such behavior, always discard the last output argument, or don't request scalar outputs from size() that way.

% a 384x512x3 RGB image
A = imread('peppers.png');

% scalar outputs will mislead you unless you know what they mean
[h,w] = size(A)

h = 384

w = 1536

% if you only need the height and width, request and discard the last output
[h,w,~] = size(A)

h = 384

w = 512

I prefer to use vector outputs, but that requires attention as well.

% sz is a variable-length vector, so it's hard to use programmatically
sz = size(A)

sz = 1x3

384 512 3

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% in newer versions, you can explicitly get the dimensions you need all at once
% that way the vector length is always fixed
sz = size(A,1:2)

sz = 1x2

384 512

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