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How can I change the LSBs of consecutive elements of an array?

Asked by fiona rozario on 2 Jun 2017
Latest activity Commented on by fiona rozario on 3 Jun 2017
I have a row matrix a = [23, 255, 67, 89, 45, 90, 34, 12] and a number, n = 3 (of 8 bits) which I want to hide in the LSBs of every element. 3 is '00000011', so LSB of 23 should change to 0 (if not already 0), LSB of 255 should change to 0 and so on till 12. n is a variable from 2-8 of 8 bits.
I tried the following code:
n=3;
a=[23, 255, 67, 89, 45, 90, 34, 12];
str=dec2bin(n,8);
for i=1:8
b(1,i)=bitset(a(1,i),8,str(i));
end
I get the following error: ASSUMEDTYPE must be an integer type name.
How do I convert the string 'str' to be a numeric array of the 1s and 0s?

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2 Answers

Answer by Joseph Cheng
on 2 Jun 2017
Edited by Joseph Cheng
on 2 Jun 2017

you could use bitand()
a = [23, 255, 67, 89, 45, 90, 34, 12]';
n=3;
b = bitand(a ,255-n,'uint8')
disp(dec2bin(a))
disp('zeroed out last 2 bits')
disp(dec2bin(b))
i know i didn't implement your selection of n correctly as i wasn't able to follow what you were describing with lsbs of 3 and the 23 example, however the bit wise operators should work whatever you're trying to describe

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n is a user input, so suppose n = 3, in binary - '00000011' (stored in 'str'). a(1) = 23 or '00010111'.
What I want to do is change the least significant bit of a(1) from 1 to whatever is the first bit of n; in this case - '0'. So, a(1) in binary should change to '00010110' (22 in decimal). Similarly, the least significant bit of 255 should be changed to whatever is the second bit of n....and so on till the eighth element of the array, a.

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Answer by Guillaume
on 2 Jun 2017

The third argument of bitset must either be the number 0 or 1 or a char array representing a valid type (such as 'uint8'). You're giving it the char '0' or '1'. Change your line to:
b(1, i) = bitset(a(1, i), 8, str(i) - '0'); %or str(i) == '1', whichever you prefer

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