Hex 01A9 to 16-binary?

I am trying to convert this input 01A9 into binary

4bit Hex input

01A9

16bit Binary Output

0000 0001 1010 1001

**no spaces

Answer 1

[Walter Roberson]

H = '01A9';
dec2bin(sscanf(H, '%4x'),16)

ans = '0000000110101001'

Note that the result will be character.

H can have more than 4 characters, and will be interpreted 4 characters at a time from the beginning . So if H is not an exact multiple of 4 characters, this particular algorithm will not implicitly pad from the begining with 0's -- only the last group will be 0 padded.

dec2bin(sscanf('01A9876', '%4x'),16)

ans = 2x16 char array
    '0000000110101001'
    '0000100001110110'

This was interpretered as 01A9 (0)876.

If you needed to be padded with leading 0s implicitly then that could certainly be programmed.

Answer 2

[Stephen23]

No loops, no padding, works correctly for any number of hex digits (not just 16 bits):

H = '01A9';
B = reshape(dec2bin(sscanf(H,'%1x'),4).',1,[])

B = '0000000110101001'

Answer 3

[James Tursa]

E.g., if H is not too big:

H = '01A9';
B = dec2bin(hex2dec(H),numel(H)*4)

Otherwise you will need some form of a loop (explicit or hidden in a function call). E.g.,

C = arrayfun(@(h)dec2bin(hex2dec(h),4),H,'uni',false);
B = [C{:}]