C++ get each digit in int

Question

I have an integer:

int iNums = 12476;

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

foreach(iNum          


        

Answer 1

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 );
}

Answer 2

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 3

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 4

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 5

void print_each_digit(int x)
{
    if(x >= 10)
       print_each_digit(x / 10);

    int digit = x % 10;

    std::cout << digit << '\n';
}

Answer 6

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;
    }
}