my_atoi

//1、找到数字字符或者加减号//2、计算值时需要保证不溢出，将要溢出时返回最大或最小值int myatoi(const char *str)
{
    if ( NULL == str )
    {
        printf("error, param is NULL\n");
        return 0;
    }

    int nIndex = 0;

    while ( ‘\0‘ != *(str + nIndex) )
    {
        if ( (*(str + nIndex) >= ‘0‘ && *(str + nIndex) <= ‘9‘)
           || *(str + nIndex) == ‘-‘
           || *(str + nIndex) == ‘+‘ )
        {
            break;
        }

        ++nIndex;
    }

    if ( ‘\0‘ == *(str + nIndex) )
    {
        printf("warn, there is no dest \n");
        return 0;
    }

    // is negative number ?
    bool nFlag = false;

    if ( *(str + nIndex) == ‘-‘ )
    {
        nFlag = true;
        ++nIndex;
    }
    else if ( *(str + nIndex) == ‘+‘ )
    {
        ++nIndex;
    }

    int nRet = 0;

    while ( ‘\0‘ != *(str + nIndex) && *(str + nIndex) >= ‘0‘ && *(str + nIndex) <= ‘9‘ )
    {
        if ( nRet > INT_MAX / 10 )
        {
            return INT_MAX;
        }
        else if ( nRet < INT_MIN / 10 )
        {
            return INT_MIN;
        }
        else
        {
            if ( nRet == INT_MAX / 10 && *(str + nIndex) >= ‘8‘ )
            {
                return INT_MAX;
            }
            else if ( nRet == INT_MIN / 10 && *(str + nIndex) == ‘9‘ )
            {
                return INT_MIN;
            }
        }

        int  temp = *(str + nIndex) - ‘0‘;
        if ( nFlag )
        {
            temp *= -1;
        }

        nRet =  nRet*10 + temp;

        ++nIndex;
    }

    return nRet;
}