How to separate first 8 digits in binary format and convert them to the hex?

I need a code to ask me to enter a number in binary format, and then separate them byte by byte (each 8 digits) then displace low and high value bytes and convert them to the hex.

for example if the input is 1000000011000000 then the output would be c080

Answer 1

[dpb]

Enter the value as a string and then convert. First below does the endian swap directly; if want to do this way write a little helper function that does it.

Alternatively, use the builtin swapbytes function on the internal representation.

MATL
>> b='1000000011000000'
b =
1000000011000000
>> dec2hex(bin2dec([b(9:16) b(1:8)]))
ans =
C080
>> dec2hex(swapbytes(uint16(bin2dec(b))))
ans =
C080
>>

Answer 2

[Image Analyst]

Try this:

binaryString = '1000000011000000'
decimalNumber = bin2dec(binaryString)
hexString = dec2hex(decimalNumber)

Adapt as necessary. To get substrings

ss = s(1:8) % Get first 8 digits.

To reverse digits:

sr = s(end: -1 : 1)

Answer 3

[Jan]

bin2dec and dec2hex have a certain overhead, which can be avoided:

str = '1000000011000000';
bin = reshape(str - '0', 8, [])';
d   = bin * pow2(:-1:0)';
d   = d([2,1]);  % Swap bytes
h   = sprintf('%x', d);

[Mike]

Thank you Jan.