How to divide 8 bit binary into two 4 bit?

For example A=11001111,

i want to break it as A1=1100 and A2=1111.

How it can be done in matlab?

Answer 1

[madhan ravi]

 A = '11001111'
 A = mat2cell(A, 1, [4, 4]);
 celldisp(A)

[madhan ravi]

Or:

 A = num2cell(reshape(A, 4, []).', 2);
 celldisp(A)

[sami ullah]

If i change A to like this

A = '1100'

A = mat2cell(A, 1, [4, 4]);

celldisp(A)

it gives the following error

Error using mat2cell (line 97)

Input arguments, D1 through D2, must sum to each dimension of the input matrix size, [1 4].'

I need output like this

A1=0000

A2=1100

[sami ullah]

I did it as:

N=12;

A=dec2bin(A,8)

A=mat2cell(A, 1, [4, 4]);

celldisp(A)

And it is done

Thanks

Answer 2

[Xavier]

I think it would be more efficient in a bit-wise operation performing the modulus and the division.

Start with an 8-bit binary number. For example, let's use the binary number 11011010. To split this number into two 4-bit groups, we need to find the remainder of the number when divided by 16 (which is 2^4, the number of bits in a 4-bit group).

To do this, we can use the modulus operator (%), which gives us the remainder after division.

So, 11011010 % 16 = 10.

The remainder we obtained in step 2 represents the least significant 4 bits of the original number. To get the most significant 4 bits, we need to divide the original number by 16 and discard the remainder. We can use integer division to do this.

So, 11011010 / 16 = 1101.

We now have two 4-bit groups: 1101 and 1010. The first group (1101) represents the most significant 4 bits of the original number, and the second group (1010) represents the least significant 4 bits.

Therefore, the original 8-bit binary number 11011010 can be split into two 4-bit groups: 1101 and 1010.

[Walter Roberson]

Note that integer division for this purpose is idivide

If you were to use a uint8() / 16 then the output uint8 would be rounded.

There are other approaches. For example,

A = randi([0 255], 5, 1, 'uint8')

A = 5×1

89 241 191 120 243

top_bits = bitand(A, 240) / 16

top_bits = 5×1

5 15 11 7 15

bottom_bits = bitand(A, 15)

bottom_bits = 5×1

9 1 15 8 3

%cross check
dec2hex(A, 2) == [dec2hex(top_bits,1), dec2hex(bottom_bits,1)]

ans = 5×2 logical array
   1   1
   1   1
   1   1
   1   1
   1   1