Problem with a function: to convert from base3 to base10

[dpb]

No idea what you did to get the flashing display when you posted, but just attach the text of the window and any error messages...this is unreadable.

Answer 1

[DGM]

For the future, please use the formatting tools available to you to paste formatted code. Don't post pictures of code. Nobody can run a picture of code, least of all a flashing picture of code that mysteriously disappears and is replaced by an incorrectly rotated error message.

Your function works at a basic level, but it has problems. It does not guard against invalid inputs (e.g. 546 is not a valid base-3 number). Avoid using while loops to blindly chew at numbers if you can precalculate the number of iterations beforehand. My guess is that you fed the function a negative number and the while loop just kept going until it ate all the available memory.

Long story short: sanitize your inputs and don't use while loops unnecessarily. The whole thing simplifies anyway.

num3=min(max(num3,0),2);
x=num3;
% this will get the number of digits for reasonably sized integers
nt=numel(num2str(fix(abs(num3))));
num10=0;
for a=1:nt
	num10=num10+mod(x,10)*3^(a-1);
	x=floor(x/10);
end

[base2dec(num2str(num3),3) num10]

If you don't need to make this function as an exercise or challenge, you can always use base2dec().

a=base2dec('22',3);

gives

a =
     8

[Teresa Perdicoulis]

I am soory, but with num3=121 it does not give the right result.

[Teresa Perdicoulis]

May be this:

num3=max(num3,0);

if (num3 > 2)

x=num3;

% this will get the number of digits for reasonably sized integers

nt=numel(num2str(fix(abs(num3))));

num10=0;

for a=1:nt

num10=num10+mod(x,10)*3^(a-1);

x=floor(x/10);

end

else num10 = num3;

end

[Teresa Perdicoulis]

Thank you.