warning: format '%s' expects argument of type 'char *', but argument 2 has type 'int' [closed]

Question

I have this program that claims to convert a decimal number to hexadecimal. The issue is that the program throws the above warning and I don't know how to correct it. The return (variable identifier) is supposed to be a %s because what you want to display is a string.

#include <stdio.h>

int main() {
    int n = 0;
    char hex = '\0';
    const char * HEX_DIG = "0123456789ABCDEF";
    printf ("Enter a positive integer: ");
    scanf ("%d",&n);
    do{
        hex = HEX_DIG [n % 16] + hex;
        n = (int) n/16;
    }while ( n != 0 );
    printf ("\nHexadecimal= %s", hex);
    return 0;
}