atoi 函数实现

 1  int atoi(const char *str)  
 2     { 
 3         if (str == nullptr) return 0; 
 4         //注意：判断空字符串数组 
 5         if(*str == ‘\0‘) return 0; 
 6         long long llnum = 0; 
 7         int sign = 1; 
 8    
 9         while (*str == ‘ ‘) 
10         { 
11             str++; 
12         } 
13         if (*str == ‘-‘) 
14         { 
15             sign = -1; 
16             str++; 
17         } 
18         //注意：不要溜了要判断+号的情况 
19         else if(*str == ‘+‘) 
20         { 
21             str++; 
22         } 
23         while (*str >= ‘0‘ && *str <= ‘9‘) 
24         { 
25             int i = *str - ‘0‘; 
26             llnum = llnum*10 + i; 
27             if (llnum*sign > INT_MAX) 
28             { 
29                 llnum = INT_MAX; 
30                 break; 
31             } 
32             else if (llnum*sign < INT_MIN) 
33             { 
34                 llnum = INT_MIN; 
35                 break; 
36             } 
37             str++; 
38         } 
39         return int(sign*llnum); 
40     }