Split an Integer into its digits c++

Question

I\'m trying to learn c++ on my own and I\'ve hit a bit of a road block. The problem is I need to take an integer,split it into its digits and get the sum of the digits and

Answer 1

I used this for input for 5 numbers

int main () {

using namespace std;

int number;`

cout << "Enter your digit: ";
cin >> number;
if (number == 0)
  return 0;

int a = number % 10;
int second = (number / 10);

int b = second % 10;
int third = (second / 10);

int c = third % 10;
int fourth = (third / 10);

int d = fourth % 10;
int fifth = (fourth / 10);

int e = fifth % 10;

cout << e << " " << d << " " << c << " " << b << " " << a << endl;

return 0;

}

Answer 2

Push the digits onto a stack.

After you've gotten all the digits,

 sum = 0 ;
 while( stack not empty ) {
   pop the stack to get a digit
   sum += digit
   display digit
 }
 display sum

You need a stack, you say? Use the STL's stack: std::stack<int>

Answer 3

Your problem comes from the fact that you are reading the digits backwards, thus you need to print them out backwards. A stack will help you tremendously.

#include "stdafx.h"
#include <cstdlib>
#include <iostream>
#include <math.h>
#include <stack>

int countDigitsInInteger(int n)
{
    int count =0;
    while(n>0)
    {
        count++;
        n=n/10;
    }
    return count;
}

using namespace std;

int main(int argc, char *argv[])
{  
    int intLength =0;
    int number;
    int digit;      
    int sum = 0;
    string s;    
    cout << "Please enter an integer ";
    cin >>number;
    cout << "Orginal Number = "<<number <<endl;
    //make the number positive
    if (number<0)
        number = -number;    

    intLength = countDigitsInInteger(number);
    //break apart the integer into digits

    stack<int> digitstack;
    while(number>0)
    {                         
        digit = number % 10;
        number = number / 10;
        digitstack.push(digit);
        sum = sum+digit; 
    }

    while(digitstack.size() > 0)
    {
        cout << digitstack.top() << " ";
        digitstack.pop();
    }

    cout <<endl <<"Sum of the digits is: "<<sum<<endl;
    system("PAUSE");
    return EXIT_SUCCESS;
}

Oh, and BTW, keep your indentation clean. Its important.

EDIT: In response to Steve Townsend, this method is not necessarily overkill, it is just different from yours. The code can be slimmed down so that it seems less like overkill:

#include <iostream>
#include <stack>
#include <string>

using namespace std;

int getInput(string prompt)
{
    int val;
    cout << prompt;
    cin >> val;
    return val < 0 ? -val : val;
}

int main(int argc, char** argv)
{
    int num = getInput("Enter a number: ");
    cout << "Original Number: " << num << endl;

    stack<int> digits;
    int sum = 0;
    while(num > 0)
    {
        digits.push(num % 10);
        sum += digits.top();
        num = num / 10;
    }

    while(digits.size() > 0)
    {
        cout << digits.top() << " ";
        digits.pop();
    }

    cout << endl << "Sum of digits is " << sum << endl;

    return 0;
}

Answer 4

Use std::stack to store separate digits, then print out the contents of stack.

here is a nice article: http://en.wikipedia.org/wiki/Stack_(data_structure)

here is how to use stacks in C++: http://www.sgi.com/tech/stl/stack.html

Answer 5

Let's not forget the stringstream approach, which I also find elegant.

#include <iostream>
#include <sstream>

int main()
{
    int num = 123456789;
    std::cout << "Number: " << num << std::endl;

    std::stringstream tmp_stream;
    tmp_stream << num;
    std::cout << "As string: " << tmp_stream.str() << std::endl;

    std::cout << "Total digits: " << tmp_stream.str().size() << std::endl;


    int i;
    for (i = 0; i < tmp_stream.str().size(); i++)
    {
        std::cout << "Digit [" << i << "] is: " << tmp_stream.str().at(i) << std::endl;
    }

    return 0;
}

Outputs:

Number: 123456789
As string: 123456789
Total digits: 9
Digit [0] is: 1
Digit [1] is: 2
Digit [2] is: 3
Digit [3] is: 4
Digit [4] is: 5
Digit [5] is: 6
Digit [6] is: 7
Digit [7] is: 8
Digit [8] is: 9

Answer 6

A simple solution:

int n = 12345;
vector<int> digits;
while (n != 0) {
    digits.insert(digits.begin(), n%10);
    n /= 10;
}
for(auto & i : digits)
    cout << i << " ";

output: 1 2 3 4 5