ex-3-5.c

/*
 * Exercise 3-5. Write the function  itob(n,s,b) that converts the integer n 
 * into a base  b character representation in the string  s. In particular,  
 * itob(n,s,16) formats  s as a hexadecimal integer in s.  
 */

#include <stdlib.h>
#include <stdio.h>
#include <limits.h>

#define abs(x) ((x) > 0) ? (x) : -(x)
#define MAX_LEN 50

void itob(int n, char s[], int b);
void reverse(char s[]);

int main(void) {
    char buffer[MAX_LEN];
    int i;
    
    for ( i = 2; i <= 36; ++i ) {
        itob(INT_MIN, buffer, i);
        printf("Decimal %d in base %-2d : %s\n", INT_MIN, i, buffer);
    }
    return 0;
}
 
 
/*  Stores a string representation of integer n
    in s[], using a numerical base of b. Will handle
    up to base-36 before we run out of digits to use.  */
 
void itob(int n, char s[], int b) {
    static char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    int i, sign;
    
    if ( b < 2 || b > 36 ) {
        fprintf(stderr, "EX3_5: Cannot support base %d\n", b);
        exit(EXIT_FAILURE);
    }
    
    sign = n;
    i = 0;
    do {
        s[i++] = digits[abs(n % b)];
    } while ((n /= b) != 0);
    if (sign < 0)
        s[i++] = '-';
    s[i] = '\0';
    reverse(s);
}
 
 
/*  Reverses string s[] in place  */
 
void reverse(char s[]) {
    int c, i, j;
    for ( i = 0, j = strlen(s)-1; i < j; i++, j--) {
        c = s[i];
        s[i] = s[j];
        s[j] = c;
    }
}