Answer to Exercise 4-12, page 88
Solution by Gregory Pietsch
Adapt the ideas of printd to write a recursive version of atoi ; that is, convert an integer into a string by calling a recursive routine.
/*
itoa() is non-standard, but defined on p.64 as having this prototype:
void itoa(int n, char s[])
Instead of this, I thought I'd use a different prototype (one I got from
the library manual of one of my compilers) since it includes all of the
above:
char *itoa(int value, char *digits, int base);
Description: The itoa() function converts an integer value into an
ASCII string of digits. The base argument specifies the number base for
the conversion. The base must be a value in the range [2..36], where 2
is binary, 8 is octal, 10 is decimal, and 16 is hexadecimal. The buffer
pointed to by digits must be large enough to hold the ASCII string of
digits plus a terminating null character. The maximum amount of buffer
space used is the precision of an int in bits + 2 (one for the sign and
one for the terminating null).
Returns: digits, or NULL if error.
*/
#include <stdlib.h>
char *utoa(unsigned value, char *digits, int base)
{
char *s, *p;
s = "0123456789abcdefghijklmnopqrstuvwxyz"; /* don't care if s is in
* read-only memory
*/
if (base == 0)
base = 10;
if (digits == NULL || base < 2 || base > 36)
return NULL;
if (value < (unsigned) base) {
digits[0] = s[value];
digits[1] = '\0';
} else {
for (p = utoa(value / ((unsigned)base), digits, base);
*p;
p++);
utoa( value % ((unsigned)base), p, base);
}
return digits;
}
char *itoa(int value, char *digits, int base)
{
char *d;
unsigned u; /* assume unsigned is big enough to hold all the
* unsigned values -x could possibly be -- don't
* know how well this assumption holds on the
* DeathStation 9000, so beware of nasal demons
*/
d = digits;
if (base == 0)
base = 10;
if (digits == NULL || base < 2 || base > 36)
return NULL;
if (value < 0) {
*d++ = '-';
u = -value;
} else
u = value;
utoa(u, d, base);
return digits;
}
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