itoa函数的实现

void itoa(int n, char *s)

{

  int sign;

  char *t = s;

  

  if ((sign = n) < 0)

    n = -n;

  

  do

  {

    *s++ = n % 10 + '0';

  } 

  while ((n /= 10) >0);

  

  if (sign < 0)

    *s++ = '-';

  

  *s = '\0';

  

  reverse(t);

}

void reverse(char *s)

{

  int c;

  char *t;

  

  for (t = s + (strlen(s) - 1); s < t; s++, t-- )

  {

    c = *s;

    *s = *t;

    *t = c;

  }

}

原文来自：http://blog.21ic.com/user1/5433/archives/2008/54208.html

本文转自 Linux_woniu 51CTO博客，原文链接:http://blog.51cto.com/linuxcgi/1965340