I want to write a recursive Function that can calculate the sum of all the digit. If the input is 12345 then the answer will be 1+2+3+4+5 , without using string2num and loop.

function output= digit_sum(n)
if n<10
output=fix(n);
end
if n>0
output=output+digit_sum(n*0.1);
end
end
I wrote this code but the problem I am facing is if i set output=0; anywhere then in all the recalling function process my result will turn to be 0, How to Solve this ?

4 Comments

When n>=10 then you do not define output before that line, so when you refer to
... output + ...
MATLAB has no idea what output should be, because you have not defined it.
You need to consider your logic so that output is defined for all values before you try to use it.
function output= digit_sum(n)
output=0;
if n<10
output=fix(n);
end
if n>0
output=digit_sum(n*0.1);
end
end
If I set output=0 anywhere then the result is coming 0 .
Maybe you should take a step back and write down the algorithm on paper using words instead of code. Then take a simple 2 digit example and run it through your algorithm on paper. Once you have things figured out for your algorithm, then turn it into code and start testing. That will force you to understand the algorithm first, before you even get to the coding stage.

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 Accepted Answer

function output= digit_sum(n,a)
if nargin==1
a=0;
end
if n<10
output=a+fix(n);
else
output=digit_sum(floor(n*.1),a+mod(n,10));
end
end

3 Comments

There is just one input number digit_sum(n) .I just tried this but the output says 'Unrecognized FUnction or variable 'output'' Where i am mistaking ?
function output= digit_sum(n)
if n>0
output= (n/10 -fix(n/10))*10+digit_sum(fix(n*0.1));
end
end
Using nargin, your function can have 1 or more inputs.
output = digit_sum(12345);
Try the above on my function.
function output= digit_sum(n)
if n<10
output=fix(n);
end
if n>0
output=((n/10 -fix(n/10))*10)+digit_sum(fix(n*0.1));
end
end
Thanks , I got the problem where i was mistaking,Now the code runs fine. Thanks for your help.
function output= digit_sum(n)
if n<10
output=fix(n);
end
if n>0
output=mod(n,10)+digit_sum(fix(n*0.1));
end
end

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More Answers (7)

You can do this with any inbuit functions.
function x = digit_sum(n)
x=0;
if n>0
x=mod(n,10)+digit_sum(floor(n./10));%recursive
end
function x=digit_sum(inp)
if inp<10
x=inp;
else
x = mod(inp,10) + digit_sum(fix(inp/10));
end
end
function output= digit_sum(n) if n<10 output=fix(n); end if n>0 output=output+digit_sum(n*0.1); end end
function sum = digit_sum(num)
if fix(num)==0
sum=0;
else
x = fix(num/10);
sum = rem(num,10) + digit_sum(x);
end
end
%is this code correct.output is right but i am not able to understand when
%if fix(num)==0 condition will true than value of sum should be zero. but
% zero is not output ,why?
function out=digit_sum(in)
q=in;
a=q/10;
b=floor(a);
c=q-b*10;
w=c;
if q>0
w=w+digit_sum(b);
end
out=w;
end
Hi, i have just done this homework. Here is my code:
function output = digit_sum(A)
remA = rem(A,10); % calculates the last digit of A
if isequal(A,0)
output = 0;
else
A = (A-rem(A,10))/10; % gives the new A with substracting the last digit
output = remA + digit_sum(A);
end
end
function x=digit_sum(n)
x=0;
if n<=0
x=0;
else
x=x+mod(n,10)+digit_sum((n-mod(n,10))/10);
end
end

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Asked:

on 17 Aug 2020

Answered:

on 25 Oct 2023

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