Alternative method to dec2bin function

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

[Jan]

This sounds like a homework question. Then please post, what you have tried so far and ask a specific question.

Hint: To get the first bit, use:

x = 17;
rem(x, 2)

To get the others, you can divide by 2 successively and round.

[Haseeb Hashim]

I have tried similar to this but we have to divide the resulting quotient form first division. You have given the function which only gives us the remiander. ineed to automate the code.

[Haseeb Hashim]

Oh I get it now, thanks for help man

I think we should use the floor function instead of round to get 3 if answer is 3.5

[Jan]

This is the floor() function. Another hint: You do not need loops:

x = 17;
y = x ./ [1, 2, 4, 8, 16]

With floor() you get the natural numbers and rem() extracts the first bit. You can express [1,2,4,8] as pow2(0:3).

[Haseeb Hashim]

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]

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.

Answer 1

[Jan]

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))