Displaying sum of digits without loops

[Sam Clark]

Ok so here is my issue. I need to display the sum of digits in a year without using loops. For example if the input is 1986 then the sum is 24.

I know that with loops I would do this:

int YearSum(0);
int x(1);
for (x = 1; x <= 4; ++x)
{
YearSum = YearSum + Year % 10;
Year = Year / 10;
}

But what about without loops? I've tried Year % 10 + Year / 10 which gets me 204... I'm lost without loops...

[Rabbit]

To get the first digit, it would be the year integer division by 1000. The second digit would be the year integer division by 100 mod 10. And so forth.

[weaknessforcats]

OK. Set Result to year. Result is 1966.
Now Calculate Thousands = Result / 1000 gets the 1. Your first dight.
Now calculate Result = Result - Thousands*1000
which is 1966 - 1000 leaves you 966.
Now calculate Hundreds = Result / 100 gets the 9. Your second digit
Now calculate Result = Result - Hundreds *100
etc...

You could also << to a stringstream and then >> to a string. Now you would have string[3] == '1' so the int value would be string[3] - 48 to get from ASCII to decimal. string[0] would be '9', etc...

There's probably a recursive function solution where each recursion returns one of the values. You would recurse down until you reached 4 levels and then return summing the values as you come up.

You might also ask someone else for the answer and just write it down. No loop there. :)

[donbock]

I'm aware of three ways to implement repetitive actions such as this: iteration (looping); recursion; and tediously sequential (loop unrolling). All of these have been mentioned already.

[whodgson]

weaknessforcats last suggestion is possibly the most effective solution but you could also do this:

int a,b,c,d,sum;
cout<<"\nenter the year (with spaces)\n";
cin>>a>>b>>c>>d;
cout<<"the year was "<<a<<b<<c<<d;
sum=a+b+c+d;
cout<<"\nthe sum of year digits: "<<sum<<endl;