无人驾驶入门--Autoware使用手册
无人驾驶汽车系统入门 - Autoware_UsersManual_v1.1Autoware-用于城市自主驾驶的集成开源软件,支持以下功能:3D本地化3D映射路径规划路径跟随加速/制动/转向控制数据记录汽车/行人/物体检测交通信号检测交通灯识别车道检测对象跟踪传感器校准传感器融合面向云的地图连接自动化智能手机导航软件仿真虚拟现实Autoware受BSD许可证保护。请自行负责使用。为了安全使用,我们为不拥有真正自主车辆的人提供基于ROSBAG的仿真方法。如果您使用Autoware与真正的自主车辆,请在现场测试之前制定安全措施和风险评估。工尹-刁一定義亡机飞求世。Chapter1。 About This documentThis chapter describes the purpose of this document.Autoware Users ManualAutoware Developers manualChapter2。 RoS and AutowareBefore operating Autoware, RoS and Autoware are described in this chapter.Robot middleware- rosecently, the wide range potential of robotics has been focused by not only roboticsexperts but also non-robotics experts to join robot development It is believed thatthis trend leads robotics to advancement and developments to other domainsHowever, robot development is getting harder because the advancement and the complexityof robot functions have been increased Unlike Pcs and smartphones robotic developmenthas considered various hardware, OS, programming languages. Hence, the differences havemainly been obstructed for robotic developers as well as robotics experts to join robotdevelopmentTo solve the problem, the demand of making common platforms has been increased, andsome platforms have been published Within a common platform, developers can combinevarious software published by other developers, and speed up development by reusing themTherefore, it is expected that developers can more focus on fields of interestROS (Robot Operating System) is a framework for robotic software development. It wasdeveloped by Willow Garage in U.S. OSFR (Open Source Robotics Foundation) has currentlymaintained it. The open sourced ros has mainly been used by communities in U.S. andEurope as well as communities in JapanNote that ros has "os" in its name however, it is not "os" like Windows or linux. It is amiddleware that runs on uniX based OSROS FeaturesOriginal build system(Catkin)mage processing library(OpenCvv Data logging tool (ROSBAG)v Visualization tools for data and software state(rvizCoordinate transformation library (TFQt based gul development tool (RQTA""file is used to start multiple nodes atAutowareutoware is open source software based on ros. autoware is pushed on github forautonomous driving research and development. Most of autonomous drivingsystem consist of recognition, judgment, and operation. Autoware providesnecessary functions, such as 3-D map generation, localization, object recognition, andvehicle control, for autonomous drivingROS PCAutoware〔Ros)RecognitionJudgmentoperationObject DetectionLane KeepingAcceleratorLocalizationIntersectionBrakeSteeringAutowareRiderAutoware routePath PlanningVehicleVehicle control(CAN)CFigure 1 Autoware overviewwhile support systems such as driving assistance and safetydiagnosis support, use multi-core CPU3-D Map Generation and SharingLocalization(NDT: Normal Distributions Transform)object DetectionPath GenerationAutonomous Drivingtarget speed. In addition, the route includes landmarks, "way point, setintervals. The autonomous driving system operates path following by following the wayUser InterfaceA user interface called " Runtime Manager ofpath following. Furthermore, a tablet user interface, "Autoware Rider, of Autoware enablesROS PCAutoware(ROS)Runtime ManagerVIZAutoware ridertabletDisplay on vehicleOculusFigure 2 User interfacePlatform structure for autowareApplicationAutowareMiddlewareROSHost OsLinux( UbuntuCPUGPUCameraGNSSLIDARFigure 3 Platform structure for Autoware
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2017最全华为机试题C/C++(含答案源码)
2017最全华为机试题C/C++(含答案源码),包含111道上机考试题,欢迎下载,觉得资源好请好评。分别将字符串中的字符转换成整型数字,进行计算后,再转换成字符类型存储起来数为其中和是输入,是的长度,是的长度。是输出4.删除子串,只要是原串中有相同的子串就删掉,不管有多少个,返回子串个数输出删除后的字符串删除子串5.约瑟夫环是一个数学的应用问题:已知n个人(以编号1,2,3..n分别表示)围坐在一张圆桌周围。从编号为k的人开始报数,数到m的那个人出列:他的下一个人又从1开始报数,数到m的那个人又出列;依此规律重复下去,直到圆桌周围的人仝部出列。6.比较一个数组的元素是否为回文数组比较两个数组,要求从数组最后一个元素廾始逐个元素冋前比较,如果2个数组长度不等,则只比较较短长度数组个数元素。请编程实现上述比较,并返回比较中发现的不相等元素的个数比如:数组{1,3,5}和数组77,21,1,3,5}按题述要求比较,不相等元素个数为0数组{1,3,5}和数组:77,21,1,3,5,7按题述要求比较,不相等元素个数为3要求实现函数int array compare(int len1, int array1[], int len2, int array2[l输入】 int len1:输入被比较数组1的元素个数;int array l[]:输入被比较数组1;int lcn2:输入被比较数组2的元素个数;int array2L]:输入被比较数组2【输出】无【返回】不相等元素的个数,类型为int小例1)02: int array1[ =11,3, 5, int len1=3, int array 2=77, 21, 1, 3, 51int e函数返回:02)输入: int array1[]=:1,3,5),int1en1=3, int array2={7,21,1,3,5,7int lend6函数返回:3约瑟大环变种:输入一个由随机数组成的数列(数列中每个数均是大于0的整数,长度已知),和初始计数值m。从数列首位置开始计数,计数到m后,将数列该位置数值替换计数值m,并将数列该位置数值出列,然后从下一位置从新开始计数,直到数列所有数值出列为止。如果计数到达数列尾段,则返回数列首位置继续计数。请编程实现上述计数过程,同时输出数值岀列的顺序比如:输入的随机数列为:3,1,2,4,初始计数值m-7,从数列首位置开始计数(数值3所在位置)第一轮计数出列数字为2,计数值更新m2,出列后数列为3,1,4,从数值4所在位置从新开始计数第二轮计数出列数字为3,计数值更新m3,出列后数列为1,4,从数值1所在位置开始计数第三轮计数出列数字为1,计数值更新m=1,出列后数列为4,从数值4所在位置开始计数最后一轮计数出列数字为4,计数过程完成。输出数值出列顺序为:2,3,1,4。要求实现函数id array iterate(int len, int input array [, int m, int output array [)输入】 int len:输入数列的长度;int Intput array[]:输入的初始数列intm:初始计数值【输出】 int output array[]:输出的数值出列顺序【返回】无示例输入: int input array[13,1,2,4}, int lcn4输出: output array[]2,3,1,4手机弓码合法性:问题描述:我国大陆运营商的手机号码标准格式为:国家码+手机号何,例如:8613912345678。特点如下:、长度13位2、以86的国家码打头3、手机号码的每一位都是数字。请实现手机号码合法性判断的函数要求1)如果手机号码合法,返回02)如果手机号码长度不合法,返回13)如果于机号码中包含非数字的字符,返回24)如果于机号码不是以86打头的,返回3:【注】除成功的情况外,以上其他合法性判断的优先级依次降低。也就是说,如果判断出长度不合法,直接返回1即可,不需要再做其他合法性判断。要求实现函数int verifyMsisdn (chark inMsisdn)【输入】char* inmsisdn,表示输入的手机号码字符串。【输出】无【返回】判断的结果,类型为int示例输入: inMsisdn=“869123456789“输出:无返回:1输入: msisdn=“88139123456789输出:无输入: inMsisdn=“86139123456789“输出:无返简单的四则运算问题描述:输入一个只包含个位数字的简单四则运算表达式字符串,计算该表达式的值注:1、表达式只含,,(,),四则运算符2、表达式数值只包含个位整数(0-9),且不会出现0作为除数的情况3、要考虑加减乘除按通常四则运算规定的计算优先级4、除法用整数除法,即仅保留除法运算结果的整数部分。比如8/3=2。输入表达式保证无0作为除数情况发生5、输入字符串一定是符合题意合法的表达式,其屮只包括数字字符和四则运算符字符,除此之外不含其它任何字符,不会出现计算溢出情况要求实现函数:int calculatc(int lcn, char *cxpStr输入】 int cn:字符串长度;char* cxpStr:表达式字符串【输出】无【返回】计算结果示例1)输入:char* expstr“1+4*5-8/3函数返回:192)输入:char* expStr=“8/3*3”函数返回:6
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