自写atoi实现函数

闲着没事，写了一下atoi函数的实现。

atoi函数的实现主要有三个方面：

1、字符的过滤原则

2、符号的判定

3、int类型的溢出问题。

这是实现代码：#include <iostream>#include <assert.h> using namespace std;const char* filteZero( const char *pStr ){assert( pStr );const char *p = pStr;while( *p != '/0' && *p == '0' )p++;return p;}void filteSpace( const char *&pStr ){assert( pStr );while ( *pStr != '/0' && *pStr ==' ' )pStr++;}int getSign( const char* &pStr ){if( *pStr == '-' && *pStr != '/0' ){pStr = pStr + 1;return -1;}else{ if( *pStr == '+')pStr ++;return 1;}}int _atoi(const char *pStr ){assert(pStr);int rlt = 0;enum _state{ safe, check, overflow } state = safe;filteSpace(pStr);int sign = getSign( pStr );pStr = filteZero( pStr );while ( *pStr <= '9' && *pStr >= '0'){if ( state == check ){if ( sign > 0 && *pStr > '7' ){rlt = INT_MAX; }else if ( sign < 0 && *pStr > '8' ){rlt = INT_MIN;}else rlt = 10 * rlt + (*pStr - '1') + 1; break;}if ( state == overflow ){rlt = sign > 0 ? INT_MAX : INT_MIN;break;}rlt = 10 * rlt + (*pStr - '1') + 1; pStr ++;if ( sign > 0 ){if ( rlt < INT_MAX / 10 ) state = safe;else if ( rlt = INT_MAX / 10 )state = check;elsestate = overflow;}if ( sign < 0 ){if ( rlt > INT_MIN / 10 )state = safe;else if ( rlt == INT_MIN /10 )state = check;elsestate = overflow;}}return rlt * sign;}void main(){char *p = "-21474837880s";assert( atoi(p) == _atoi(p) );cout << _atoi(p) << endl;system("pause");}

如果有bug，或优化的办法，等待发现。