I used the following code for hiding a binarystring inside an image using lsb using the following code but an error occurred.

Question

Anushka 2015-8-2

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/232038-i-used-the-following-code-for-hiding-a-binarystring-inside-an-image-using-lsb-using-the-following-co

回答： Image Analyst 2015-8-2

I used the following code for hiding a binary string inside an image using lsb using the following code but an error occurred.

   k = 1;
   for r=1:3
   for i = 1 : m
       for j = 1 : n
       C = dec2binvec(double(c(i,j,r)),8); %convert decimal number to binary vector
       if k <= ml                          %embedd till message length
              C(8) = binaryString(k);          %embedding in LSB
           end                            %convert binary vector to decimal number
           s(i,j,r) = binvec2dec(C);
           k = k + 1;
       end
   end    
   end

the error is Attempted to access c(1,385,1); index out of bounds because size(c)=[256,384,3].

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2015-8-2

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/232038-i-used-the-following-code-for-hiding-a-binarystring-inside-an-image-using-lsb-using-the-following-co#answer_187961

在 MATLAB Online 中打开

You do not show us how you determined the value of m and n.

I suspect that you used

[m, n] = size(c);

If I am correct then you need to read the description of size() again:

n < ndims(X)  di equals the size of the ith dimension of X for 0<i<n, but dn equals the product of the sizes of the remaining dimensions of X, that is, dimensions n through ndims(X).

Now follow that for the case of having two size() outputs for a 3D array.

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Anushka 2015-8-2

编辑：Image Analyst 2015-8-2

在 MATLAB Online 中打开

Yes I used

[m,n]=size(c);

What might be the reason for the error such that I didn't get it correctly?

请先登录，再进行评论。

Answer 2

Image Analyst 2015-8-2

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/232038-i-used-the-following-code-for-hiding-a-binarystring-inside-an-image-using-lsb-using-the-following-co#answer_187970

在 MATLAB Online 中打开

You say size(c)=[256,384,3] but then you say you got the sizes of c by doing this:

[m,n]=size(c);

Never use size that way with images. Since you say that c is a color image with 3 color channels, then n is really the number of columns multiplied by the number of color channels. Why? See Steve's blog: http://blogs.mathworks.com/steve/2011/03/22/too-much-information-about-the-size-function/. To correct, do this:

[m, n, numberOfColorChannels] = size(c);

or even better because it's more explicit and descriptive:

[rows, columns, numberOfColorChannels] = size(c);

Then rename m to rows and n to columns in the rest of your code.

I don't know what dec2binvec() function does, but make sure that it returns a single uint8 number because that's what

s(i,j,r) = binvec2dec(C);

requires. Finally, change

for r=1:3

to

for colorChannel = 1 : numberOfColorChannels

and rename "r" to the much more descriptive, understandable, and maintainable "colorChannel". Why numberOfColorChannels instead of 3? Because in general, the images might be grayscale and not have 3 planes. This will make it more general and robust and it will still work for color images.