Write a function (Not a built-in function) that converts a given number n in base 2 to base 10

Suppose n is a base2 number, which will be put in as a vector (i.e., 10100 = [1 0 1 0 0]). I'm trying to write a code than converts that number to a base10 number. I ran it and got the following error: Out of memory. The likely cause is an infinite recursion within the program.

Error in base2 (line 4) y(i) = base2(i)*(2^(length(n)-i));

What am I doing wrong? The code is:

    function [base10] = base2(n)
y = [];
for i = 1:length(n)
    y(i) = base2(i)*2^(length(n)-i);
    base10 = sum(y);
end
end

[Stephen23]

https://www.mathworks.com/matlabcentral/fileexchange/58850-base2base-nstart-bstart-bend-

Answer 1

[Guillaume]

Learn to use the debugger. Step through your code line by line, look at how the variables change and where the code go at each step. You'll quickly see where it goes wrong.

You're using recursion (your function calls itself), but there's nothing in your code to stop the recursion. Look at your loop, the first step is to call base2 again.

Once you've fixed that, a more minor issue is that you recalculate the final sum in the loop, when you only need to do it once, when the loop is finished.

Answer 2

[George]

There's a builtin function called bin2dec which turns a binary string into a decimal number. So if you convert the data into a string you can use bin2dec.

    aa = [1 0 1 0];
    bb = num2str(aa);
    result = bin2dec(bb)

   result =

    10

[Guillaume]

This is homework, they're not supposed to use built-in function.

Answer 3

[Thorsten]

There is a nice oneliner that solves the problem:

base10 = 2.^(numel(n)-1:-1:0)*n(:);

If you use it, your task is to understand it, in case your advisor asks...