C++ get each digit in int

I have an integer:

int iNums = 12476; 

And now I want to get each digit from iNums as integer. Something like:

foreach(iNum in iNums){ printf("%i-", iNum); } 

So the output would be: "1-2-4-7-6-". But i actually need each digit as int not as char.

Thanks for help.

17 Answers

void print_each_digit(int x) { if(x >= 10) print_each_digit(x / 10); int digit = x % 10; std::cout << digit << '\n'; } 

Convert it to string, then iterate over the characters. For the conversion you may use std::ostringstream, e.g.:

int iNums = 12476; std::ostringstream os; os << iNums; std::string digits = os.str(); 

Btw the generally used term (for what you call "number") is "digit" - please use it, as it makes the title of your post much more understandable :-)

Here is a more generic though recursive solution that yields a vector of digits:

void collect_digits(std::vector<int>& digits, unsigned long num) { if (num > 9) { collect_digits(digits, num / 10); } digits.push_back(num % 10); } 

Being that there are is a relatively small number of digits, the recursion is neatly bounded.

I don't test it just write what is in my head. excuse for any syntax error

Here is online ideone demo

vector <int> v; int i = .... while(i != 0 ){ cout << i%10 << " - "; // reverse order v.push_back(i%10); i = i/10; } cout << endl; for(int i=v.size()-1; i>=0; i--){ cout << v[i] << " - "; // linear } 

Here is the way to perform this action, but by this you will get in reverse order.

int num; short temp = 0; cin>>num; while(num!=0){ temp = num%10; //here you will get its element one by one but in reverse order //you can perform your action here. num /= 10; } 

You can do it with this function:

void printDigits(int number) { if (number < 0) { // Handling negative number printf('-'); number *= -1; } if (number == 0) { // Handling zero printf('0'); } while (number > 0) { // Printing the number printf("%d-", number % 10); number /= 10; } } 

To get digit at "pos" position (starting at position 1 as Least Significant Digit (LSD)):

digit = (int)(number/pow(10,(pos-1))) % 10; 

Example: number = 57820 --> pos = 4 --> digit = 7

To sequentially get digits:

int num_digits = floor( log10(abs(number?number:1)) + 1 ); for(; num_digits; num_digits--, number/=10) { std::cout << number % 10 << " "; } 

Example: number = 57820 --> output: 0 2 8 7 5

Drawn from D.Shawley's answer, can go a bit further to completely answer by outputing the result:

void stream_digits(std::ostream& output, int num, const std::string& delimiter = "") { if (num) { stream_digits(output, num/10, delimiter); output << static_cast<char>('0' + (num % 10)) << delimiter; } } void splitDigits() { int num = 12476; stream_digits(std::cout, num, "-"); std::cout << std::endl; } 

I don't know if this is faster or slower or worthless, but this would be an alternative:

int iNums = 12476; string numString; stringstream ss; ss << iNums; numString = ss.str(); for (int i = 0; i < numString.length(); i++) { int myInt = static_cast<int>(numString[i] - '0'); // '0' = 48 printf("%i-", myInt); } 

I point this out as iNums alludes to possibly being user input, and if the user input was a string in the first place you wouldn't need to go through the hassle of converting the int to a string.

(to_string could be used in c++11)

I know this is an old post, but all of these answers were unacceptable to me, so I wrote my own!

My purpose was for rendering a number to a screen, hence the function names.

void RenderNumber(int to_print) { if (to_print < 0) { RenderMinusSign() RenderNumber(-to_print); } else { int digits = 1; // Assume if 0 is entered we want to print 0 (i.e. minimum of 1 digit) int max = 10; while (to_print >= max) // find how many digits the number is { max *= 10; digits ++; } for (int i = 0; i < digits; i++) // loop through each digit { max /= 10; int num = to_print / max; // isolate first digit to_print -= num * max; // subtract first digit from number RenderDigit(num); } } } 

Based on @Abyx's answer, but uses div so that only 1 division is done per digit.

#include <cstdlib> #include <iostream> void print_each_digit(int x) { div_t q = div(x, 10); if (q.quot) print_each_digit(q.quot); std::cout << q.rem << '-'; } int main() { print_each_digit(12476); std::cout << std::endl; return 0; } 

Output:

1-2-4-7-6- 

N.B. Only works for non-negative ints.

My solution:

void getSumDigits(int n) { std::vector<int> int_to_vec; while(n>0) { int_to_vec.push_back(n%10); n=n/10; } int sum; for(int i=0;i<int_to_vec.size();i++) { sum+=int_to_vec.at(i); } std::cout << sum << ' '; } 

The answer I've used is this simple function:

int getDigit(int n, int position) { return (n%(int)pow(10, position) - (n % (int)pow(10, position-1))) / (int)pow(10, position-1); } 

Hope someone finds this helpful!

You can do it using a while loop and the modulo operators. It just gives the digits in the reveser order.

int main() { int iNums = 12476; int iNum = 0; while(iNums > 0) { iNum = iNums % 10; cout << iNum; iNums = iNums / 10; } } 

// Online C++ compiler to run C++ program online

#include <iostream> #include <cmath> int main() { int iNums = 123458; 

// int iNumsSize = 5;

int iNumsSize = trunc(log10(iNums)) + 1; // Find length of int value for (int i=iNumsSize-1; i>=0; i--) { int y = pow(10, i); // The pow() function returns the result of the first argument raised to the power of the second argument. int z = iNums/y; int x2 = iNums / (y * 10); printf("%d ",z - x2*10 ); // Print Values } return 0; } 

int a; cout << "Enter a number: "; cin >> a; while (a > 0) { cout << a % 10 << endl; a = a / 10; } 

int iNums = 12345; int iNumsSize = 5; for (int i=iNumsSize-1; i>=0; i--) { int y = pow(10, i); int z = iNums/y; int x2 = iNums / (y * 10); printf("%d-",z - x2*10 ); }