How to convert decimal into binary?

109 visualizaciones (últimos 30 días)
Sky Scrapper
Sky Scrapper el 22 de En. de 2019
Editada: Jan el 1 de Nov. de 2021
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?
  4 comentarios
Jan
Jan el 22 de En. de 2019
2^8 or 2^8-1 ?
Stephen23
Stephen23 el 22 de En. de 2019
Editada: Stephen23 el 22 de En. de 2019
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

Iniciar sesión para comentar.

Respuesta aceptada

Jan
Jan el 22 de En. de 2019
Editada: Jan el 1 de Nov. de 2021
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.
  9 comentarios
Walter Roberson
Walter Roberson el 23 de En. de 2019
A one-million bit binary number cannot be converted to a double precision value.
Jan
Jan el 23 de En. de 2019
If
dec2bin(0:2^n-1, 8) - '0'
is working, calling
Dec2BinNumeric(0:2^n-1, 8)
is not a serious difference.

Iniciar sesión para comentar.

Más respuestas (2)

PRAVEEN GUPTA
PRAVEEN GUPTA el 8 de Jul. de 2019
i have string of number [240 25 32 32]
i want to convert them in binary
how can i do this???
  2 comentarios
Jan
Jan el 8 de Jul. de 2019
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
AB WAHEED LONE el 6 de Mzo. de 2021
Editada: AB WAHEED LONE el 6 de Mzo. de 2021
I know it is late but somwhow it may help
bin_array=dec2bin(array,8)-'0';

Iniciar sesión para comentar.


vandana Ananthagiri
vandana Ananthagiri el 5 de Feb. de 2020
function A = binary_numbers(n)
A = double(dec2bin(0:((2^n)-1),n))-48;
end
  2 comentarios
Walter Roberson
Walter Roberson el 5 de Feb. de 2020
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
vincent voogt el 1 de Nov. de 2021
Maybe a late reply, but dec2bin return as string of ASCII characters, where 0-9 are mapped on character number 48-57.

Iniciar sesión para comentar.

Categorías

Más información sobre Characters and Strings en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by