How to convert decimal into binary?

Hello,

I need to convert n-bit decimal into 2^n bit binary number. I do not have much idea. Can anybody help me please?

[Jan]

What exactly is a "n bit decimal"? Integer or floating point values? What about dec2bin?

[Sky Scrapper]

It is integer. I have tried:

n= 8;

for i = 0:2^n-1

x = dec2bin(i,8);

end

It's showing, x= 11111111

But I need the values of x=0.......2^8 in binary

[Jan]

2^8 or 2^8-1 ?

[Stephen23]

Get rid of the loop:

>> V = 0:pow2(8)-1;
>> dec2bin(V)
ans =
00000000
00000001
00000010
00000011
00000100
00000101
00000110
00000111
00001000
00001001
00001010
... lots of rows here
11111010
11111011
11111100
11111101
11111110
11111111

Answer 1

[Jan]

This code shows '11111111' only, because you overwrite the output in each iteration:

n= 8;
for i = 0:2^n-1
   x = dec2bin(i,8);
end

Therefore x contains the last value only: dec2bin(2^n-1, 8).

Better:

x = dec2bin(0:2^n-1, 8);

Or if you really want a loop:

n = 8;
x = repmat(' ', 2^n-1, 8);  % Pre-allocate
for i = 0:2^n-1
   x(i+1, :) = dec2bin(i,8);
end
x

[EDITED] If you want the numbers 0 and 1 instead of a char matrix with '0' and '1', either subtract '0':

x = dec2bin(0:2^n-1, 8) - '0';

But to avoid the conversion to a char and back again, you can write an easy function also:

function B = Dec2BinNumeric(D, N)
  B = rem(floor(D(:) ./ bitshift(1, N-1:-1:0)), 2);
end
% [EDITED] pow2(n) reülaced by faster bitshift(1, n)

PS. You see, the underlying maths is not complicated.

[Sky Scrapper]

yes, that's working. Thank you so much.

[Sky Scrapper]

Problem is that it's showing string value say, ''11111111'' but i will have to get double array something like ''1 1 1 1 1 1 1 1''. i think it's not possible using dec2bin. I am using Matlab2016a. so it's not possible to use ''decimalToBinaryVector''. colud you please help me know in this regard?

[Jan]

To convert from '1010' to [1 0 1 0], see [EDITED] in my answer.

[Sky Scrapper]

If i run your function code it's showing error,''Not enough input arguments.''

[Walter Roberson]

What arguments did you pass to Dec2BinNumeric ?

[Jan]

Call it e.g. like:

B = Dec2BinNumeric(17, 8)

[Sky Scrapper]

i will have to convert for ''n'' having higher values as like n=1000000.

Anyway, ''x = dec2bin(0:2^n-1, 8) - '0';'' this is working properly. thanks!

[Walter Roberson]

A one-million bit binary number cannot be converted to a double precision value.

[Jan]

If

dec2bin(0:2^n-1, 8) - '0'

is working, calling

Dec2BinNumeric(0:2^n-1, 8)

is not a serious difference.

Answer 2

[PRAVEEN GUPTA]

i have string of number [240 25 32 32]

i want to convert them in binary

how can i do this???

[Jan]

Do no attach a new question as an asnwer of another one.

Did you read this thread? dec2bin has been suggested already, as well as a hand made solution Dec2BinNumeric. Simply use them.

[AB WAHEED LONE]

I know it is late but somwhow it may help

bin_array=dec2bin(array,8)-'0';

Answer 3

[vandana Ananthagiri]

function A = binary_numbers(n)
A = double(dec2bin(0:((2^n)-1),n))-48;
end

[Walter Roberson]

Why 48?

I know the answer, but other people reading your code might not, so I would recommend either a comment or a different representation.

[vincent voogt]

Maybe a late reply, but dec2bin return as string of ASCII characters, where 0-9 are mapped on character number 48-57.