I have this so far:
data = 14
out = dec2bin(data, 4)
which gives:
out = 1110
But I want to get binary number in this format:
out = [1 1 1 0]
Thanks for help!
Add a comment
|
3 Answers
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
1 Comment
Matjaž
Yes but this require to use communications package and I can't use it
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
P.S. More common answer provided here: dec2base with independent bits/digits calculation
Comments
Yet another way: Use "bitget":
data = 14
out = bitget (data, 4:-1:1)
out =
1 1 1 0
