datahex = sprintf('%02x', T1)??

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Radwa
Radwa on 16 Jan 2015
Commented: Radwa on 16 Jan 2015
T1 =00111001 10001010 01011001 1011010010101100000000000100100011010001010110011110001001101010111100110111101111
datahex =
30303131313030312031303030313031302030313031313030312031303131303130303130313031313030303030303030303030313030313030303131303130303031303130313130303131313130303031303031313031303130313131313030313130313131313031313131
T1=[T1 count]
datahex = sprintf('%02x', T1)
I don't know why output in that format],T1 is character

Accepted Answer

Guillaume
Guillaume on 16 Jan 2015
Edited: Guillaume on 16 Jan 2015
First, if there are spaces in your T1 string, get rid of them:
T1(T1 == ' ') = []; %remove spaces
For demo purpose:
T1 = char(randi(double(['0' '1']), 1, 128)); %random demo string of 128 '0' and '1'
Then, assuming that the length of T1 is a multiple of 8:
datahex = num2cell(lower(dec2hex(bin2dec(reshape(T1, [], 8)))), 2)'
Basically, reshape your string into columns of 8 characters. Convert each row from binary string to decimal number and then from decimal number to hexadecimal string. Optionally, convert to lowercase and cell array.
  3 Comments
Guillaume
Guillaume on 16 Jan 2015
Doh! Sorry, should have tested and thought through the code better.
Reshape into 8 rows and transpose, rather than 8 columns:
datahex = num2cell(lower(dec2hex(bin2dec(reshape(T1, 8, [])'))), 2)'
Radwa
Radwa on 16 Jan 2015
thanks so much, I have another Question If I want to add 1 first to the 128 bit then convert, the problem I face is that to add 1 it must be dec, and if I use bin2dec it must be less than 52 bit

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More Answers (1)

Geoff Hayes
Geoff Hayes on 16 Jan 2015
Edited: Geoff Hayes on 16 Jan 2015
Radwa - your T1 is a string of characters, so the zeros and ones are characters. The ASCII code for a character 0 and character 1 is 48 and 49 respectively. The hexadecimal equivalent of each is 30 and 31, so your above answer makes sense. (The 29 would be for the space character.)
  2 Comments
Radwa
Radwa on 16 Jan 2015
ّI want to deal with it as binary string and convert to hexadecimal string
I count deal with bin2hex for 2reasons: 1. count will be 128 bits 2. I want to make it in this format {'39' '8a' '59' 'b4' 'ac' '00' '00' '00''00' '00' '00' '00' '00' '00' '00' '00'}
Geoff Hayes
Geoff Hayes on 16 Jan 2015
I don't understand your format. How does your input string translate to that? There are limitations to the size of integers that you can use with these functions, and if you insist on using 128 bits, you may need to create a class that can handle a data type of that size.

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