C库函数——字符串转数字整理

atof(将字符串转换成浮点型数)
atoi(将字符串转换成整型数)
atol(将字符串转换成长整型数)
strtod(将字符串转换成浮点数)
strtol(将字符串转换成长整型数)
strtoul(将字符串转换成无符号长整型数)
toascii(将整型数转换成合法的ASCII 码字符)
toupper(将小写字母转换成大写字母)
tolower(将大写字母转换成小写字母)

 int atoi(const char *nptr)
 {
         int c;              /* current char */
         int total;         /* current total */
         int sign;           /* if '-', then negative, otherwise positive */

         /* skip whitespace */
         while ( isspace((int)(unsigned char)*nptr) )
             ++nptr;

         c = (int)(unsigned char)*nptr++;
         sign = c;           /* save sign indication */
         if (c == '-' || c == '+')
             c = (int)(unsigned char)*nptr++;    /* skip sign */

         total = ;

         while (isdigit(c)) {
             total =  * total + (c - '');     /* accumulate digit */
             c = (int)(unsigned char)*nptr++;    /* get next char */
         }

         if (sign == '-')
             return -total;
         else
             return total;   /* return result, negated if necessary */
 }

atoi函数源码

 double atof(char s[])   //将字符串s转换成double型的浮点数
 {  

     int i;
     int sign;
     int flag;
     int expn;
     double val;
     double power;  

     sign = ;
     flag = ;
     power = 1.0;
     expn = ;
     for(i = ; isspace(s[i]); ++i)
         ;
     if(s[i] == '-')
         sign = -;
     if(s[i] == '+' || s[i] == '-')
         ++i;
     for(val = 0.0; isdigit(s[i]); ++i)
         val = val * 10.0 + (s[i] - '');
     if(s[i] == '.')
         ++i;
     for(; isdigit(s[i]); ++i)
     {
         val = val * 10.0 + (s[i] - '');
         //power = power * 10.0;
         ++flag;
     }
     if(s[i] == 'e' || s[i] == 'E')  //如果写成s[i++] == 'e' || s[i++] == 'E'，if(s[i] == '-')
                                     //则当输入的字符串带有E时，不能正确得到结果，这是因为在一
                                     //个语句中使用两次自增操作，引起歧义
         if(s[++i] == '-')
         {
             ++i;
             for(; isdigit(s[i]); ++i)
                 expn = expn *  + (s[i] - '');
             expn = expn + flag;
             power = pow(, expn);
             return sign * val / power;
         }
         else
         {
             for(; isdigit(s[i]); ++i)
                 expn = expn *  + (s[i] - '');
             expn = expn - flag;
             power = pow(, expn);
             return sign * val * power;
         }  

     power = pow(, flag);
     return sign * val / power;  

 }  

atof函数代码