C++ get each digit in int - c++

Question

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.

Answer 1

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

Answer 2

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 :-)

Answer 3

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.

Answer 4

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;
}

Answer 5

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
}

Answer 6

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

Answer 7

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;
}
}

Answer 8

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;
}

Answer 9

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)

Answer 10

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);
}
}
}

Answer 11

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.

Answer 12

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 << ' ';
}

Answer 13

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!

Answer 14

// 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;
}

Answer 15

You can do it using a while loop and the modulo operators.
It just gives the digits in the revese order.
int main() {
int iNums = 12476;
int iNum = 0;
while(iNums > 0) {
iNum = iNums % 10;
cout << iNum;
iNums = iNums / 10;
}
}

Answer 16

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

Answer 17

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 );
}