# How can i replace 0 by -1 in all binary number till 256 as 00000000, 00000001, 00000010, 00000011, 00000100, ..........​​.........​.​........​..​...., 11111111 ?

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kailash kumar on 17 Jul 2015
Commented: arich82 on 17 Jul 2015
Without using loops
Guillaume on 17 Jul 2015
If you replace the 0s by 1 in all the binary numbers, then the result is 255 for all of them.

Azzi Abdelmalek on 17 Jul 2015
Edited: Azzi Abdelmalek on 17 Jul 2015
a=dec2bin(0:255)
out=strrep(cellstr(a),'0','1')
Azzi Abdelmalek on 17 Jul 2015
a=dec2bin(0:255)
out=strrep(cellstr(a),'0','-1')

arich82 on 17 Jul 2015
Edited: arich82 on 17 Jul 2015
[Edited to use 0:255 instead of 1:256]
n = 8;
a = dec2bin(0:n-1) - '0'
returns a as a matrix of doubles:
a =
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Now just use a logical mask to change 0 --> -1:
a(a == 0) = -1
which returns
a =
-1 -1 -1
-1 -1 1
-1 1 -1
-1 1 1
1 -1 -1
1 -1 1
1 1 -1
1 1 1
For your case, use n = 256. The complete solution (I think) is:
n = 256;
a = dec2bin(0:n-1) - '0';
a(a==0) = -1;
disp(a);
Please accept this answer if it helps, or let me know in the comments if it's not what you're looking for.
arich82 on 17 Jul 2015
Alternative to dec2bin: bitget.
Unfortunately, it seems bitget can only get either several bits from a scalar, or a single bit from each element of an array. Wrapping it in bsxfun gets around this apparent limitation:
a = bsxfun(@bitget, [0:255].', 8:-1:1);
a(a == 0) = -1;
Not an issue for your use case, but for much larger arrays, I'd expect bitget to be much quicker than all of the character conversions going on in the dec2bin version, though I could be wrong. (You'd also want to use nextpow2 to get all digits for the more general case).
n = 256;
a = bsxfun(@bitget, (0:n - 1).', nextpow2(n - 1):-1:1);
a(a == 0) = -1;
There's also de2bi if you have the appropriate toolbox (Communications System Toolbox?):