1 /*  2 atoi算法，要求完美版  3 有两种，一种是用longlong，一种是真用int  4 “”  5 "  "  6 “-”  7 “+”  8 “  -23”  9 “  +23” 10 “12a" 11 "abc" 12 越界情况 13 */ 14 #include <iostream> 15 using namespace std; 16  17 //方法一用 long long 取巧 18 int atoi(const char * str,bool & flag) 19 { 20     long long result=0; 21     int tmp; 22     int sign=1; 23     if(str==NULL) 24     { 25         flag=false; 26         return 0; 27     } 28     while(isspace(*str)) 29         str++; 30     //if(*str==‘\0‘) 31     //{ 32     //    flag=false; 33     //    return 0; 34     //} 35     if(*str==‘+‘||*str==‘-‘) 36     { 37         if(*str==‘-‘) 38             sign=-1; 39         str++; 40     } 41     if(*str==‘\0‘)                     //”    “和”+“ ，”-“的情况 42     { 43         flag=false; 44         return 0; 45     } 46     while(*str>=‘0‘&&*str<=‘9‘) 47     { 48         tmp=*str-‘0‘; 49         result=result*10+tmp; 50         if(sign==-1) 51         { 52             if(result-1>INT_MAX)           //这里注意，如果是INT_MAX+1，系统或默认转换成int，导致越界，所以用result-1，result是longlong不会越界 53             { 54                 result=INT_MAX+1; 55                 flag=false; 56                 break; 57             } 58         } 59         else if(result>INT_MAX) 60         { 61             result=INT_MAX; 62             flag=false; 63             break; 64         } 65         str++; 66     } 67     if(*str==‘\0‘) 68         flag=true; 69     else 70         flag=false; 71     return sign*result; 72 } 73  74 //方法二 用int 75 int atoi2(const char * str ,bool & flag) 76 { 77     int result=0; 78     int sign=1; 79     int tmp; 80     int MAX_DIV=INT_MAX/10; 81     int MAX_R=INT_MAX%10; 82     int MIN_R=MAX_R+1;   //或者为MIN_R=-(INT_MIN%10);    //如果用(INT_MAX+1)%10就错了，直接加的时候越界 83     int MIN_DIV=(INT_MAX)/10;         84  85     //july用的是max=(int)((unsigned)~0>>1); 86     //min_r=(int)(((unsigned)~0>>1+1)%10); 87     if(str==NULL) 88     { 89         flag=false; 90         return result; 91     } 92     while(isspace(*str)) 93         str++; 94     if(*str==‘+‘||*str==‘-‘) 95     { 96         if(*str==‘-‘) 97             sign=-1; 98         str++; 99     }100     if(*str==‘\0‘)101     {102         flag=false;103         return result;104     }105     while(*str>=‘0‘&&*str<=‘9‘)106     {107         int tmp=*str-‘0‘;108         if(sign==1)109         {110             if(result>MAX_DIV||(result==MAX_DIV&&tmp>MAX_R))111             {112                 result=INT_MAX;113                 flag=false;114                 return result;115             }116         }117         else if(result>MIN_DIV||(result==MIN_DIV&&tmp>MIN_R))118         {119             result=INT_MIN;                                               //result直接为最小-2147483648，直接返回了120             flag=false;121             return result;122         }123         result=result*10+tmp;124         str++;125     }126     if(*str==‘\0‘)127         flag=true;128     else129         flag=false;130     return sign*result;131 }132 133 134 135 int main()136 {137     char * str[]={""," ","abc","  123","  -1234","   +34","  12bc","+","-","+++","0","2147483647","+2147483648","-2147483648","-2147483649","+12123131312432342","   -122323434534534654"};138     //cout<<INT_MAX<<endl;139     int n=17;140     bool flag;141     for(int i=0;i<n;i++)142     {143         int tmp=atoi2(str[i],flag);144         if(tmp==0)145         {146             if(flag==false)147                 cout<<str[i]<<" 错误数据转换为： "<<tmp<<endl;148             else149                 cout<<str[i]<<" 转换为: "<<tmp<<endl;150         }151         else152         {153         cout<<str[i]<<" 转换为: "<<tmp<<endl;154         }155     }156     system("pause");157 }