Decimal to binary as double type array, not string

Question

I have this so far:

data = 14
out = dec2bin(data, 4)

which gives:

out = 1110

But I want to get binary num

Answer 1

You're looking for de2bi, bi2de functions.

It requires the Communication Systems Toolbox.

If you don't have it, you can work define these functions at the begining of your code as:

de2bi = @(x) dec2bin(x)>48;
bi2de = @(x) x*2.^(size(x,2)-1:-1:0)';

Test:

dec = 1:10
bin = de2bi(dec)
dec = bi2de(bin)

Output:

dec =

    1    2    3    4    5    6    7    8    9   10

bin =

  0  0  0  1
  0  0  1  0
  0  0  1  1
  0  1  0  0
  0  1  0  1
  0  1  1  0
  0  1  1  1
  1  0  0  0
  1  0  0  1
  1  0  1  0

dec =

    1
    2
    3
    4
    5
    6
    7
    8
    9
   10

P.S. If, by some reason, you don't want to use dec2bin at all, you can define de2bi function as:

In Matlab/Octave:

de2bi = @(x)  2.^[(floor(log2(max(x(:)))):-1:1),0];
de2bi = @(x) rem(x(:),2*de2bi(x))>(de2bi(x)-1);

In Octave only (as far as Octave allows default values for anonymous functions):

By default returns the same as in the previous example, but optional bit position parameter is available:

de2bi = @(dec,bit=[(1+floor(log2(max(dec(:))))):-1:2, 1]) rem(dec(:),2.^bit)>(2.^(bit-1)-1);

%Call examples:
x=0:10; 
n=3:-1:1;
de2bi(x)
de2bi(x,n)   % returns only the three least significant bits

Answer 2

You're looking for de2bi with the 'left-msb' option.

data = 14
out = de2bi(data, 4,'left-msb')

Which requires the Communication Systems Toolbox though. Alternatively use your original approach with the fundamental dec2bin with the following addition:

data = 14
out = double( dec2bin(data, 4) ) - 48

---

out =

     1     1     1     0

Answer 3

Yet another way: Use "bitget":

data = 14
out = bitget (data, 4:-1:1)
out =
   1   1   1   0