Printing each digit of an integer

[mikey108]

Task:
1. In main() funtion, ask user to enter some positive number. If number is not positive, terminate the program.
2. Otherwise you call a seperate function from main() that takes number and:
-prints digits of this number;
-returns/prints the sum of digits

#include <iostream>
using namespace std;

int both(int n) {

    int remainder;
    int sum = 0;

    while (n>0) {
        remainder = n%10;
        sum = sum + remainder;
        int digit = n%10;
        n = n/10;
        cout << "The digits are: " << digit << endl;
        }
    return sum;
}

int main () {

    int x;
    cout <<"Enter a number: " << endl;
    cin >> x;
        if (x<=0){
            return 0;
        }
    cout << "The sum of digits is: " << both(x) <<endl;



return 0;

}

The program should run like this:

Enter a number:
345
The digits are:
3
4
5
The sum of digits is:
12

The only part I am having trouble with is printing the digits of an integer because from my code, it prints in REVERSED order instead of ascending order. I need to use a while loop inside my function. Please help, thanks in advance!

[Hiten]

you could find how many digits you have in the first place and then work your way left to right. 1111 -> 4 digits. 1111%1000, 111 % 100 ... etc
Either that or place into an array your results and print that array in reverse.
I'm sure you'll get more answers soon :)

[Chriscpp]

To get the first digit

unsigned int leftmost_digit(unsigned int number)
{
    int divisor=1;
    int i=0;
    for(i=number;i>9;i/=10)
        divisor*=10;
    return i;
}

To get the remainder

unsigned int remainder( unsigned int number)
{
    int divisor=1;
    int i=0;
    for(i=number;i>9;i/=10)
        divisor*=10;
    return number%divisor;
}

[JLBorges]

> it prints in REVERSED

Print a digit after the more significant digits have been printed.

#include <iostream>

int primt_and_sum_digits( int number )
{
    if( number == 0 ) return 0 ;

    const int least_significant_digit = number % 10 ; // print this at the end
                         // after the more significant digits have been printed

    const int sum = primt_and_sum_digits( number / 10 ) + least_significant_digit ;

    std::cout << least_significant_digit << ", " ;

    return sum ;
}

int main()
{
    for( int v : { 12, 345, 6789, 12345678 } ) // invariant: v is positive
    {
        std::cout << "The number is " << v << ",  its digits are " ;
        const int sum = primt_and_sum_digits(v) ;
        std::cout << "  and their sum is " << sum << ".\n" ;
    }
}

http://coliru.stacked-crooked.com/a/726a9c1140fa1610

[Yanson]

keep it simple just use a string

#include <iostream>
#include <string>
using namespace std;

void print(string num);
int main()
{
	string num;

	cout << "Please enter a number: " << flush;
	getline(cin, num);

	if (num.empty() || num.front() == '-')
		return EXIT_FAILURE;
	
	print(num);

	cin.ignore(INT_MAX, '\n');
	return EXIT_SUCCESS;
}

void print(string num)
{
	cout << num;
}

the above code doesn't provide any error checking and is only meant as an example

[mikey108]

I'm supposed to use a while loop to do this though. Is this possible? :(

[JLBorges]

Use an actual stack.

#include <iostream>
#include <stack>

int primt_and_sum_digits( int number )
{
    std::stack<int> digits ;
    int sum = 0 ;

    while( number != 0 )
    {
        const int least_significant_digit = number % 10 ;

        digits.push( least_significant_digit ) ; 
        sum += least_significant_digit ;

        number /= 10 ;
    }

    // print digits from top of the stack (reverse order of push) till it is empty
    while( !digits.empty() )
    {
        std::cout << digits.top() << ", " ;
        digits.pop() ;
    }

    return sum ;
}

int main()
{
    for( int v : { 12, 345, 6789, 12345678 } ) // invariant: v is positive
    {
        std::cout << "The number is " << v << ",  its digits are " ;
        const int sum = primt_and_sum_digits(v) ;
        std::cout << "  and their sum is " << sum << ".\n" ;
    }
}

http://coliru.stacked-crooked.com/a/4ac1be1cb0848f95

[Yanson]

well you know how to reverse the number. If you reverse the number twice you get the original number.

#include <iostream>
using namespace std;

int both(int n) {

	int remainder;
	int sum = 0;

	while (n > 0)
	{
		sum = (sum * 10) + n % 10;
		n = n / 10;
	}

	n = sum;
	sum = 0;

	while (n>0) {
		remainder = n % 10;
		sum = sum + remainder;
		int digit = n % 10;
		n = n / 10;
		cout << "The digits are: " << digit << endl;
	}

	
	return sum;
}

int main() {

	int x;
	cout << "Enter a number: " << endl;
	cin >> x;
	cin.ignore();

	if (x <= 0){
		return 0;
	}
	cout << "The sum of digits is: " << both(x) << endl;


	cin.ignore();
	return 0;

}

[giblit]

The lazy approach:

#include <iostream>
#include <string>
#include <sstream>


int main()
{
    const int value = 123456789;
    std::stringstream ss;
    ss << value;
    const std::string strValue = ss.str();
    
    auto it = strValue.begin(); 
    int sum = 0;
    
    std::cout << "The digits are: ";
    while(it != strValue.end())
    {
        std::cout << *it << ' ';
        sum += *it - '0';
        ++it;
    }
    std::cout << std::endl << "The sum is: " << sum << std::endl;
}

http://coliru.stacked-crooked.com/a/1bd42a661e2b65a3

Also your professor limiting to a while loop is pointless. Anything you can do in a for loop you can do in a while loop and vice versa.

[mikey108]

Thanks to everyone for replying, especially Yanson! I now understand what I was doing wrong.