Decimal to Hex in C++

Question

I've coded this program to get a number and a base as two inputs, then prints the number in the given base, it does work well when the base is less than 10, but when I enter a base like 16, instead of printing the word "A" for 10, or F for 15, the program prints the ASCII value of the words, but I want the words themselves.

#include <iostream>
using namespace std;

int base(int n, int t);

int main()
{
    char ch;
    do {
        int n=0, t=0;
        base(n,t);
        
        cout << "\n\nDo you want to continue?(y/n)";
        cin >> ch;
    } while (ch=='y');

}

int base(int n, int t)
{
    cout << "\nPlease enter the number : ";
    cin >> n;
    cout << "\nPlease enter the base : ";
    cin >> t;
    
    int i=80;
    int x[i];
    int m=n;
    

    for (i=1; n>=t ;i++) 
    {
        int rem;
        rem = n%t;
        n = n/t;


    if (rem>9)
    { 
        switch (char(rem)) 
        {
            case 10 :
                rem = 'A';
                break;
            case 11 :
                rem = 'B';
                break;
            case 12 :
                rem = 'C';
                break;
            case 13 :
                rem = 'D';
                break;
            case 14 :
                rem = 'E';
                break;
            case 15 :
                rem = 'F';
                break;
        }
    } 
    

    x[i]=rem;
    }
    

    cout << "The number " << m << " in base " << t << " is = ";
    cout << n;
            

    for (int j=(i-1); j>0; j--)
        cout << x[j];
        

    return 0;

}

I know it's because I have declared x[i] as an integer, so that's why it's happening, but I don't know how to fix this problem.

Answer 1

To print an ASCII code as a character, you cast it to char. So, instead of

cout << x[j];

You can do

cout << static_cast<char>(x[j]);

However, you currently leave digits below 10 as a number, which after casting to char would be wrong. You could fix that by adding '0' to rem if it's less than 10.

Another possibility is changing the type of x from int to char; that way, you wouldn't have to cast.