Decimal to Hex in C-CodePudding

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.

CodePudding user response：

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.

CodePudding user response：

Here is one way to fix your code to print the letters 'A', 'B', 'C', 'D', 'E', and 'F' instead of their ASCII values.

First, instead of using a switch statement to map the integer values 10, 11, 12, 13, 14, and 15 to the corresponding letters, you can use an array of characters to do this mapping. For example:

char hex_chars[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};

Then, you can use this array to convert an integer value to the corresponding character, by using the integer value as an index into the array. For example:

int rem = n % t;
char ch = hex_chars[rem];

Here is the complete code with these changes:

#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;

    char hex_chars[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};

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

        char ch = hex_chars[rem];
        x[i] = ch;
    }

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

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

    return 0;
}

Note that I also fixed a bug in your code where you were printing n instead of m in the final output. In this code, n is the quotient obtained by dividing the original number by the base, whereas m is the original number itself.