The C Programming Language, 2nd Edition, by Kernighan and Ritchie
Exercise 3.05 on page 64

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 n as a hexadecimal integer in s .

Solution by Vidhan Gupta

/* 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 n as a hexadecimal integer in s. */

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

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

#define MAXIMUM 50

int main(){
    int b = 16;
    char s[MAXIMUM];
    itob(123456,s,b);
    printf("Base %d = %s\n",b,s);
    return 0;
}

void itob(int n, char s[], int b){
    int i = 0;
    do
    {
        if (n%b>9)
            s[i++]= n%b +'A'-10;
        else
            s[i++] = n%b +'0';
    } while ((n/=b));
    reverse(s);
}

void reverse (char s[]){
    int i , j , c;
    for (i = 0, j = strlen(s)-1; i < j; i++,j--)
    c = s[i], s[i]=s[j], s[j]=c;
}

OUTPUT:
Base 16 = 1E240

Solution by Paul Griffiths

/*

  EX3_5.C
  =======
  
  Suggested solution to Exercise 3-5
  
*/
    
#include <stdlib.h>
#include <stdio.h>
    
    void itob(int n, char s[], int b);
void reverse(char s[]);

int main(void) {
    char buffer[10];
    int i;
    
    for ( i = 2; i <= 20; ++i ) {
        itob(255, buffer, i);
        printf("Decimal 255 in base %-2d : %s\n", 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);
    }
    
    if ((sign = n) < 0)
        n = -n;
    i = 0;
    do {
        s[i++] = digits[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;
    }
}

Solution by seankndy

#include <stdio.h>

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

int main(void)
{
	int n;
	char s[100];
	
	for (int i = 2; i <= 20; i++) {
		itob(255, s, i);
		printf("decimal 255 in base %-2d : %s\n", i, s);
	}
	
	return 0;
}

void itob(int n, char s[], int b)
{
	int i, sign, r;
	
	sign = n;
	i = 0;
	do {
		r = n % b;
		if (sign < 0)
			r = -r;
		s[i++] = (r > 9 ? (r-10 + 'A') : (r + '0'));
	} while (n /= b);
	
	if (sign < 0)
		s[i++] = '-';
	s[i] = '\0';
	
	reverse(s);
}

void reverse(char s[])
{
	int i, j, t;
	
	for (j = 0; s[j] != '\0'; j++)
		;
	for (i = 0, --j; j > i; i++, j--) {
		t = s[j];
		s[j] = s[i];
		s[i] = t;
	}
}

Solution by Luke Panayi

My solution with additional functions included:

/*
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 n as a hexadecimal integer in s.
*/

/*
max base of 32
*/

#include <stdio.h>
#define BUFFER 1000

int reverse(char s[])
{
  char ch;
  int i, j;

  for(j = 0; s[j] != '\0'; j++);

  --j;

  for(i = 0; i < j; i++)
  {
    ch   = s[i];
    s[i] = s[j];
    s[j] = ch;
    --j;
  }

  return 0;
}

void itob(int n, char s[], int b) 	{
	int i, j, sign, c;
	unsigned k = 0;

	if ((sign=n) < 0)	{
		k = -n;
	}
	else	{
		k = n;
	}
	i = 0;

	do
	{
		j = k % b;
		s[i++] = (j < 9) ? (j + '0'):((j-10) + 'A');
	} while (k /= b);

	if (sign < 0)
		s[i++] = '-';
	s[i] = '\0';
	reverse(s);
}


int main()	{

	char s[BUFFER];

	itob(255, s, 32);
	printf("%s\n", s);

	return 0;
}

Solution by Miguel Degrossoli

/* 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 MAXLEN		32	/* an integer has 32-bits in my machine */
#define BASE_MIN	 2	/* well, you cannot go lower */
#define BASE_MAX	36	/* amount of alphanumeric chars available */

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

int main() {

	int n;		/* I'm getting its value from the trash */
	char s[MAXLEN + 1], b;

	printf("Value of %d in...\n", n);
	for (b = BASE_MIN; b <= BASE_MAX; b++)
	{
		itob(n, s, b);
		printf("Base %d: %s\n", b, s);
	}

	exit(0);
}

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

	int i, j, c;

	if (b < BASE_MIN || b > BASE_MAX)
	{
		printf("Valid base ranges from %d to %d.\n",
				BASE_MIN, BASE_MAX);
		exit(1);
	}

	j = 0;
	if (n == 0)
		s[j++] = '0';

	for (i = n; i > 0; i /= b)
	{
		if ((c = i % b) <= 9)
			s[j++] = c + '0';
		else
			s[j++] = c - 10 + 'A';
	}

	s[j++] = '\0';

	reverse(s);
}

void reverse(char s[]) {

	int i, n;
	char line[MAXLEN + 1];

	for (i = 0; s[i] != '\0' && i < MAXLEN; ++i)
		;

	n = 0;
	if (i == 0 || s[i] == '\0')
		line[i] = '\0';

	if (i > 0)
	{
		for (--i; i >=0; --i) {
			line[i] = s[n];
			n++;
		}
		for (i = 0; line[i] != '\0'; i++)
			s[i] = line[i];
	}
}

Sample:

miguel@Miguel-Notebook:~/Desenvolvimento/C$ ./exercise_3-5
Value of 22087 in...
Base 2: 101011001000111
Base 3: 1010022001
Base 4: 11121013
Base 5: 1201322
Base 6: 250131
Base 7: 121252
Base 8: 53107
Base 9: 33261
Base 10: 22087
Base 11: 1565A
Base 12: 10947
Base 13: A090
Base 14: 8099
Base 15: 6827
Base 16: 5647
Base 17: 4874
Base 18: 3E31
Base 19: 3439
Base 20: 2F47
Base 21: 281G
Base 22: 21DL
Base 23: 1IH7
Base 24: 1E87
Base 25: 1A8C
Base 26: 16HD
Base 27: 1381
Base 28: 104N
Base 29: Q7I
Base 30: OG7
Base 31: MUF
Base 32: LI7
Base 33: K9A
Base 34: J3L
Base 35: I12
Base 36: H1J