Separate the digits of an integer (C++) - c++

Question

I need to write a program that takes an integer as input and outputs it with its digits spaced. The algorithm works fine but prints the digits in reverse order. To get the original order, I used a for loop. The problem is that it prints nothing.
#include <iostream>
using namespace std;
int countDigit(int n)
{
int count = 0;
while (n != 0) {
n = n / 10;
++count;
}
return count;
}
int main()
{
int Num ;
int sum = 0;
int arr[100];
cout <<"Enter an integer: " ;
cin >> Num ;
int c = countDigit(Num) ;
for (int i = c; i > 0; i--)
{
arr[i] = (Num % 10);
Num = (Num / 10);
sum += arr[i];
}
cout << " Your digits:";
for (int x = 1 ; x <= c; x++)
{
cout << arr[x] << " " ;
cout << sum ;
}
return 0;
}
My issue is here
for (int x = 1 ; x <= c; x++)
{
cout << arr[x] << " " ;
cout << sum ;
}
It prints nothing. What could be the problem ?
The input used is 1234
The expected output is 1 2 3 4

Answer 1

For starters the function countDigit is wrong.
int countDigit(int n)
{
int count = 0;
while (n != 0) {
n = n / 10;
++count;
}
return count;
}
It returns 0 for the valid number 0 that has one digit.
The function can look the following way
size_t countDigit( int n, int base = 10 )
{
size_t count = 0;
do
{
++count;
} while ( n /= base );
return count;
}
In this loop
for (int i = countDigit(Num); i > 0; i--)
{
arr[i] = (Num % 10);
Num = (Num / 10);
sum += arr[i];
}
the variable Num was changed. So in the next loop
for (int x = 1 ; x <= countDigit(Num) ; x++)
{
cout << arr[x] << " " ;
cout << sum ;
}
it will not have the original value You need to use an intermediate variable to make the calculation.
Also take into account that the use can enter a negative number.
EDIT: After you updated your code then you also need to move the output of the sum from the loop
for (int x = 1 ; x <= c; x++)
{
cout << arr[x] << " " ;
}
cout << sum ;
And instead of the large integer array it is better to use an object of the type std::string.
Here is a demonstrative program
#include <iostream>
#include <string>
size_t countDigit( int n, int base = 10 )
{
size_t count = 0;
do
{
++count;
} while ( n /= base );
return count;
}
int main()
{
const int Base = 10;
int num;
std::cout << "Enter an integer: ";
std::cin >> num;
size_t count = countDigit( num );
std::string s( count, ' ' );
int sum = 0;
for ( ; count-- != 0; num /= Base )
{
int digit = num % Base;
if ( digit < 0 ) digit = -digit;
s[count] = digit + '0';
sum += digit;
}
std:: cout << "Your digits: ";
for ( const auto &c : s ) std::cout << c << ' ';
std::cout << '\n';
std::cout << "The sum of digits is " << sum << '\n';
return 0;
}
Its output might look like
Enter an integer: -123456789
Your digits: 1 2 3 4 5 6 7 8 9
The sum of digits is 45
Another approach is to write a recursive function that outputs digits and returns the sum of digits.
#include <iostream>
unsigned int recursive_sum( long long int n, long long int base = 10, std::ostream &os = std::cout )
{
long long int digit = n % base;
if ( digit < 0 ) digit = -digit;
unsigned int sum = digit + ( ( n /= base ) == 0 ? 0 : recursive_sum( n ) );
os << digit << ' ';
return sum;
}
int main()
{
int num;
std::cout << "Enter an integer: ";
std::cin >> num;
unsigned int sum = recursive_sum( num );
std::cout << '\n';
std::cout << "The sum of digits is " << sum << '\n';
return 0;
}
Again the program output might look like
Enter an integer: -123456789
1 2 3 4 5 6 7 8 9
The sum of digits is 45

Answer 2

This gives the expected output. I assigned countDigit(Num) to c and wrote the sum to outside the loop.
int main()
{
int Num ;
int sum = 0;
int arr[100];
cout <<"Enter an integer: " ;
cin >> Num ;
int c = countDigit(Num); //edit 1
// extracts the digits
for (int i = c; i > 0; i--)
{
arr[i] = (Num % 10);
Num = (Num / 10);
sum += arr[i];
}
cout << "Your digits with spaces: ";
// reverses the order
for (int x = 1 ; x <= c ; x++)
{
cout << arr[x] << " " ;
}
cout << endl << "The sum of the digits is: " << sum ; // edit 2
return 0;
}

Answer 3

Here's a shorter version:
#include <concepts>
#include <cstdio>
#include <string>
template <std::integral T>
void printDigits(T const i) {
for (auto const ch : std::to_string(i)) {
std::printf("%c ", ch);
}
std::fflush(stdout);
}
LIVE

Answer 4

Here's a take that uses strings.
#include <iostream>
#include <string>
int main()
{
std::string input;
std::size_t location;
std::cout << "Enter an integer: ";
std::getline(std::cin, input);
std::stoi(input, &location); // Don't care about the integer, just location
if (location != input.length()) {
std::cout << "Invalid Entry: Non-integer entered\n";
return 1;
}
for (auto i : input) {
std::cout << i << ' ';
}
// Here's a one liner that requires <algorithm>
// std::for_each(input.begin(), input.end(), [](auto i) { std::cout << i << ' '; });
std::cout << '\n';
}
I don't care about an integer at all. I can validate that only digits were entered and quit early if not. The only loop I need is for printing.
stoi() can return the integer, but I don't care about the integer. I just want to know how many characters were converted, and I learn that by feeding it the output argument location. Checking that location is the same as the string length ensures that only digits were entered. No decimal points, no extra letters, just numbers.
The caveats are that this might not fit into your requirements if this is an assignment. It also requires C++11, which is surprisingly not a given yet (for assignments, I get the industry arguments).