面试经典题目 atoi实现 htoi实现

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考察点：

1. 不合法检查

a. NULL

b. ""

c. "+"

d. " "

e. "124jio"

2. 溢出情况的处理

基于上述考虑的代码如下，采用一个全局状态变量来处理不合法和溢出情况。

typedef enum{INVALID, VALID} ValidFlag;ValidFlag flag;int atoi(const char *s) {   char *p;   int sign;   long long total;      if (s == NULL || *s == '\0')      return flag = INVALID;    p = s;   while (isspace((int)(unsigned int)*p))     p++;   sign = (*p == '-') ? -1 : 1;   if (*p == '+' || *p == '-'){     ++p;   }   if (*p == '\0')     // "+" "-" "  "check     return flag = INVALID;   total = 0;   while (isdigit((int)(unsigned int)*p)) {     total = total * 10 + (int)(unsigned int)*p - '0';     if (total < 0                     //overflow check          || total > (long long)INT_MAX + 1 && sign < 0         || total > (long long)INT_MAX && sign > 0)       return flag = INVALID;     p++;   }   if (*p == '\0') {        flag = VALID;     return sign == -1 ? -total : total;   }   return flag = INVALID; }

与此类似的htoi，只不过多了前缀0x或0X，以及需要处理a-f, A-F，别忘了total = total * 16 + *p - 'a' + 10;

typedef enum {INVALID, VALID} ValidFlag;ValidFlag flag;int htoi(const char *s) {  int sign;  char *p = s;  long long total = 0;  if (s == NULL || *s == '\0')    return flag = INVALID;  while(isspace(*p))    p++;  sign = (*p == '-') ? -1 : 1;  if (*p == '+' || *p == '-')    p++;  if (*p == '0' && (*(p+1) == 'X' || *(p+1) == 'x'))    p += 2;  if (*p == '\0')     //"  " "+" "0x" "-0x" check     return flag = INVALID;    while (*p) {    if (isdigit(*p))      total = total * 16 + *p - '0';    else if ('a' <= *p && *p <= 'f')      total = total * 16 + *p - 'a' + 10;     else if ('A' <= *p && *p <= 'F')      total = total * 16 + *p - 'A' + 10;     else       return flag = INVALID;            if (total < 0         || total > (long long)INT_MAX && sign > 0         || total > (long long)INT_MAX + 1 && sign < 0)      return flag = INVALID;    p++;  }    if (*p == '\0') {    flag = VALID;    return sign > 0 ? total : -total;  }  return flag = INVALID;}