ex 3-5 (the c programming language 2 ed)

itob.c

#include <stdio.h>
#include <string.h>

void copy (char to[], char from[]) {
    int i = 0;
    for (; (to[i] = from[i]) != '\0'; ++i );
}

char intChar (int n) {
    return (n >= 0 && n <= 9) ? n + 48 : ' ';
}

void reverse (char arr[]) {
    char temp[100];

    int i = 0, n = 0;

    for (i = strlen (arr) - 1; i >= 0; --i, ++n)
        temp[n] = arr[i];

    temp[n] = '\0';


    copy (arr, temp);


}

void itob (int n, char str[], int b) {
    int r = 0, x = 0, r2;

    if (b == 16) { // hexadecimal int to char conversion
        for (; n != 0; n /= 16, ++x) {
            r = n % 16;
            str[x] = (r < 10) ? r + 48 : r + 55;
            }

    } else if (b == 2) { // binary int to char conversion
        for (; n != 0; n /= 2, ++x)
            str[x] = (n % 2 == 0) ? '0' : '1';
    } else if (b == 8) { // octal int to char conversion
        for (; n != 0; ++x) {
            r2 = (n / 8) * 8;
            r = n - r2;

            str[x] = intChar(r); // converts the int into char 

            n /= 8;

        }
    }

     str[x] = '\0'; // inserting a null character to end the string 

     reverse(str); // reverses the string

}

int main ()
{
    char str[100];

    itob(670, str, 8);

    printf ("%s\n", str);

    return 0;
}