How to convert decimal into binary?

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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?
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Jan el 22 de En. de 2019
2^8 or 2^8-1 ?
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

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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.
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Walter Roberson el 23 de En. de 2019
A one-million bit binary number cannot be converted to a double precision value.
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.

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Más respuestas (2)

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???
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Jan el 8 de Jul. de 2019
Do no attach a new question as an asnwer of another one.
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';

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vandana Ananthagiri el 5 de Feb. de 2020
function A = binary_numbers(n)
A = double(dec2bin(0:((2^n)-1),n))-48;
end
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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 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.

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