在开源如此盛行的年代，那种无聊的笔试，我真是痛恶之绝！！！
可我们得无奈的面对.........

以下为经典之作，那个考官能有如此慎密思路？？不防一试！！！
-------------------------
1、自己写atoi
int atoi(number)
char  *number;
{
   int   n = 0, neg = 0;

   while (*number <= ' ' && *number > 0)
      ++number;
   if (*number == '-')
   {
      neg = 1;
      ++number;
   }
   else if (*number == '+')
      ++number;
   while (*number>='0' && *number<='9')
      n = (n * 10) + ((*number++) - '0');
   return (neg ? -n : n);
}
------------------------------------
2.
#if 1
char *
strstr(s1, s2)
const char *s1;
const char *s2;
{
   int l = strlen(s2);
   char * p = (char *) s1;

   if( l==0 ) return p;

   while (p = strchr(p, *s2))
   {
      if( memcmp(p, s2, l) == 0 )
         return p;
      p++;
   }
   return (char *) 0;
}

#else

char *strstr(str1, str2)
char *str1, *str2;
{
    char *Sptr, *Tptr;
    int len = strlen(str1) -strlen(str2) + 1;

    if (*str2)
 for (; len > 0; len--, str1++){
     if (*str1 != *str2)
  continue;

     for (Sptr = str1, Tptr = str2; *Tptr != '\0'; Sptr++, Tptr++)
  if (*Sptr != *Tptr)
      break;

     if (*Tptr == '\0')
  return (char*) str1;
 }

    return (char*)0;
}
#endif
-----------------------------
3.自己写fabs
/* Avoid using UNK if possible.  */
#ifdef UNK
#if BIGENDIAN
#define MIEEE 1
#else
#define IBMPC 1
#endif
#endif

double fabs(x)
double x;
{
union
  {
    double d;
    short i[4];
  } u;

u.d = x;
#ifdef IBMPC
    u.i[3] &= 0x7fff;
#endif
#ifdef MIEEE
    u.i[0] &= 0x7fff;
#endif
#ifdef DEC
    u.i[3] &= 0x7fff;
#endif
#ifdef UNK
if( u.d < 0 )
   u.d = -u.d;
#endif
return( u.d );
}
------------------------
4.
static char buf[12];

extern char * ultoa();

char * ltoa(val)
long val;
{
   char *p;
   int flg = 0;
   if( val < 0 ) { flg++; val= -val; }
   p = ultoa(val);
   if(flg) *--p = '-';
   return p;
}

char * ultoa(val)
unsigned long val;
{
   char *p;

   p = buf+sizeof(buf);
   *--p = '\0';

   do
   {
      *--p = '0' + val%10;
      val/=10;
   }
   while(val);
   return p;
}

--------------------
5.
#define __MAX_INT_CHARS 7

char *
itoa(i)
int   i;
{
   static char a[__MAX_INT_CHARS];
   char *b = a + sizeof(a) - 1;
   int   sign = (i < 0);

   if (sign)
      i = -i;
   *b = 0;
   do
   {
      *--b = '0' + (i % 10);
      i /= 10;
   }
   while (i);
   if (sign)
      *--b = '-';
   return b;
}

---------------------------
6：折半查
#include <stdio.h>

static int _bsearch;   /* index of element found, or where to

           * insert */

char *bsearch(key, base, num, size, cmp)
char *key;    /* item to search for */
char *base;   /* base address */
int num;      /* number of elements */
int size;    /* element size in bytes */
int (*cmp) ();   /* comparison function */
{
 int a, b, c, dir;

 a = 0;
 b = num - 1;
 while (a <= b) {
  c = (a + b) >> 1;  /* == ((a + b) / 2) */
  if ((dir = (*cmp) (key, (base + (c * size))))) {
   if (dir < 0)
    b = c - 1;
   else    /* (dir > 0) */
    a = c + 1;
  } else {
   _bsearch = c;
   return (base + (c * size));
  }
 }
 _bsearch = b;
 return (NULL);
}

太多了，一大把，哎........................