How do I split an int into its digits?

Reversed order digit extractor (eg. for 23 will be 3 and 2):

while (number > 0) { int digit = number%10; number /= 10; //print digit } 

Normal order digit extractor (eg. for 23 will be 2 and 3):

std::stack<int> sd; while (number > 0) { int digit = number%10; number /= 10; sd.push(digit); } while (!sd.empty()) { int digit = sd.top(); sd.pop(); //print digit } 

The following will do the trick

void splitNumber(std::list<int>& digits, int number) { if (0 == number) { digits.push_back(0); } else { while (number != 0) { int last = number % 10; digits.push_front(last); number = (number - last) / 10; } } } 

A simple answer to this question can be:

1. Read A Number "n" From The User.
2. Using While Loop Make Sure Its Not Zero.
3. Take modulus 10 Of The Number "n"..This Will Give You Its Last Digit.
4. Then Divide The Number "n" By 10..This Removes The Last Digit of Number "n" since in int decimal part is omitted.
5. Display Out The Number.

I Think It Will Help. I Used Simple Code Like:

#include <iostream> using namespace std; int main() {int n,r; cout<<"Enter Your Number:"; cin>>n; while(n!=0) { r=n%10; n=n/10; cout<<r; } cout<<endl; system("PAUSE"); return 0; } 

cast it to a string or char[] and loop on it

the classic trick is to use modulo 10: x%10 gives you the first digit(ie the units digit). For others, you'll need to divide first(as shown by many other posts already)

Here's a little function to get all the digits into a vector(which is what you seem to want to do):

using namespace std; vector<int> digits(int x){ vector<int> returnValue; while(x>=10){ returnValue.push_back(x%10);//take digit x=x/10; //or x/=10 if you like brevity } //don't forget the last digit! returnValue.push_back(x); return returnValue; } 

Declare an Array and store Individual digits to the array like this

int num, temp, digits = 0, s, td=1; int d[10]; cout << "Enter the Number: "; cin >> num; temp = num; do{ ++digits; temp /= 10; } while (temp); for (int i = 0; i < digits-1; i++) { td *= 10; } s = num; for (int i = 0; i < digits; i++) { d[i] = s / td %10; td /= 10; } 

int n = 1234; std::string nstr = std::to_string(n); std::cout << nstr[0]; // nstr[0] -> 1 

I think this is the easiest way.

We need to use std::to_string() function to convert our int to string so it will automatically create the array with our digits. We can access them simply using index - nstr[0] will show 1;

Start with the highest power of ten that fits into an int on your platform (for 32 bit int: 1.000.000.000) and perform an integer division by it. The result is the leftmost digit. Subtract this result multipled with the divisor from the original number, then continue the same game with the next lower power of ten and iterate until you reach 1.

You can just use a sequence of x/10.0f and std::floor operations to have "math approach". Or you can also use boost::lexical_cast(the_number) to obtain a string and then you can simply do the_string.c_str()[i] to access the individual characters (the "string approach").

I don't necessarily recommend this (it's more efficient to work with the number rather than converting it to a string), but it's easy and it works :)

#include <algorithm> #include <iostream> #include <iterator> #include <string> #include <boost/lexical_cast.hpp> int main() { int n = 23984; std::string s = boost::lexical_cast<std::string>(n); std::copy(s.begin(), s.end(), std::ostream_iterator<char>(std::cout, "\n")); return 0; } 

int n;//say 12345 string s; scanf("%d",&n); sprintf(s,"%5d",n); 

Now you can access each digit via s[0], s[1], etc

You can count how many digits you want to print first

#include <iostream> #include <cmath> using namespace std; int main(){ int number, result, counter=0, zeros; do{ cout << "Introduce un numero entero: "; cin >> number; }while (number < 0); // We count how many digits we are going print for(int i = number; i > 0; i = i/10) counter++; while(number > 0){ zeros = pow(10, counter - 1); result = number / zeros; number = number % zeros; counter--; //Muestra resultados cout << " " << result; } cout<<endl; 

}

Based on icecrime's answer I wrote this function

std::vector<int> intToDigits(int num_) { std::vector<int> ret; string iStr = to_string(num_); for (int i = iStr.size() - 1; i >= 0; --i) { int units = pow(10, i); int digit = num_ / units % 10; ret.push_back(digit); } return ret; } 

 int power(int n, int b) { int number; number = pow(n, b); return number; } void NumberOfDigits() { int n, a; printf("Eneter number \n"); scanf_s("%d", &n); int i = 0; do{ i++; } while (n / pow(10, i) > 1); printf("Number of digits is: \t %d \n", i); for (int j = i-1; j >= 0; j--) { a = n / power(10, j) % 10; printf("%d \n", a); } } int main(void) { NumberOfDigits(); } 

EASY AND BEST:

while (number > 0) { int digit = number%10; number /= 10; cout<<digit<<endl; } 

#include <iostream> using namespace std; int main() { int n1 ; cout <<"Please enter five digits number: "; cin >> n1; cout << n1 / 10000 % 10 << " "; cout << n1 / 1000 % 10 << " "; cout << n1 / 100 % 10 << " "; cout << n1 / 10 % 10 << " "; cout << n1 % 10 << " :)"; cout << endl; return 0; }