atoi函数的实现

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转自http://blog.csdn.net/udknight/article/details/1836799

int isspace(int x){ if(x==' '||x=='\t'||x=='\n'||x=='\f'||x=='\b'||x=='\r')  return 1; else   return 0;}int isdigit(int x){ if(x<='9'&&x>='0')          return 1; else  return 0;}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 = 0;        while (isdigit(c)) {            total = 10 * total + (c - '0');     /* accumulate digit */            c = (int)(unsigned char)*nptr++;    /* get next char */        }        if (sign == '-')            return -total;        else            return total;   /* return result, negated if necessary */}