itoa的实现

/* 递归与非递归版本itoa函数 */

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

#if 0
/* itoa: convert n to characters in s; recursive version */
void itoa(int n, char *s)
{
    static int i;

    if (n / 10)
        itoa(n / 10, s);
    else {
        i = 0;
        if (n < 0)
            s[i++] = '-';
    }
    s[i++] = abs(n) % 10 + '0';
    s[i] = '\0';
}
#else
void reverse(char *s)
{
#if 0
    char *head = s,
         *tail = s;

    while ('\0' != *tail)
        tail++;
    --tail;
    while (head < tail) {
        *head ^= *tail;
        *tail ^= *head;
        *head++ ^= *tail--;
    }
#else
    int c;
    char *t = NULL;

    for (t = s + (strlen(s) - 1); s < t; s++, t--) {
        c = *s;
        *s = *t;
        *t = c;
    }
#endif
}
/* itoa: convert n to characters in s; pointer version */
void itoa(int n, char *s)
{
    int sign;
    char *t = s;        /* save pointer to s */

    if ((sign = n) < 0)
        n = -n;            /* make n positive */
    do {
        *s++ = n % 10 + '0';    /* get next digit */
    } while ((n /= 10) > 0);    /* delete it */

    if (sign < 0)
        *s++ = '-';
    *s = '\0';
    reverse(t);
}
#endif

int main()
{
    int x = -1234567890;
    char buff[1024];

    itoa(x, buff);
    puts(buff);

    return 0;
}