Alternative method to dec2bin function

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I need help in the the code with alternative to bin2dec function which gives output in the double class
I have been trying but no success. Can anyone help
  6 Comments
Haseeb Hashim
Haseeb Hashim on 7 Mar 2022
Edited: Rik on 7 Mar 2022
clc
clear
close all
num = 29;
i = 1;
flag = true;
while flag == true
bin(i) = rem(num,2);
if bin(i) == 0
num = num/2;
if num == 1
bin(i+1) = 1;
break
end
elseif bin(i) ~= 0
num = floor(num/2);
if num == 1
bin(i+1) = 1;
break
end
end
i = i + 1;
end
flip(bin)
I have made this logic and it gives the exactly right answers I was wondering that is there another efficient method which gives me what want but with efficient and easy code
Rik
Rik on 7 Mar 2022
If you want to improve this code: try to determine the number of bits you need. That way you can use array operations. That removes the need for a loop and prevents dynamically growing your bin array.

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Accepted Answer

Jan
Jan on 7 Mar 2022
Edited: Jan on 7 Mar 2022
Some simplifications:
% Replace:
flag = true;
while flag == true
% by
while true
If you have checked for:
if bin(i) == 0
an
else
is enough and there is no reason to check for
elseif bin(i) ~= 0
again. But in your case there is no need to distinguish the 2 cases, because num = floor(num/2) works in all cases. Instead of catching the last bin manually, you can do this in the loop also:
num = 29;
i = 1;
while true
bin(i) = rem(num, 2);
num = floor(num / 2);
if num == 0
break
end
i = i + 1;
end
flip(bin)
Or even simpler: Check the condition instead of breaking an infinite loop:
num = 29;
bin = [];
while num > 0
bin(end+1) = rem(num, 2);
num = floor(num / 2);
end
flip(bin)
The Matlabish way is to omit the loop and to determine the number of bits in advance:
nBit = floor(log2(num)) + 1;
bin = rem(floor(num .* pow2(1-nBit:0)), 2);
Take the time to understand, what's going on here. Run it in pieces:
1-nBit:0
pow2(1-nBit:0)
num .* pow2(1-nBit:0)
floor(num .* pow2(1-nBit:0))

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