bth001复习提纲.doc

1BTH001 复习提纲 BTH001 期末考试题型：（1）判断分析题（3 题，共 6 分）判断，并简要说明（2）名词解释（3 题，共 9 分）概念简要解释（3）综合编程题（4 大题，共 85 分）每个大题下面细分若干小题包括代码判断分析、代码编写等1、类相关概念类、属性、成员函数、常量成员函数 (class/property/constant member function) 练习，1.1 理解常量成员函数（constant member function） ，在下面的 Member类中，哪些函数可以定义为常量成员函数，如何定义？1.2 实现下面类中的符号重载 operator==1.1 答案:string getName() const;2string getAddress() const;string getPhoneNr() const;string toString() const;bool operator==(const Member1.2 答案:bool Member::operator ==(const Member& member) const {return this->name == member.name }2、默认构造函数（default constructor）If no constructor is defined in a class it will get an automatically generated default constructor. Furthermore a class will automatically get agenerated- Copy constructor- Assignment operator- DestructorIf a class has member variables that are pointers it is necessary to replace the- Default constructor- Copy constructor- Assignment operator- Destructor 练习，通过默认构造函数创建对象，见 No.6 知识点练习。

Member memb;3、析构函数（Destructor）If the class has at least one member variable that is a pointer it is necessary to:- Deallocate the memory that has been allocated for the pointers by the object- 练习，见 No.16 容器类知识点练习4、复制构造函数（Copy constructor）Each class has a copy constructor. Purpose is to create an identical copy of another object.Automatically generated if not defined.If a member variable is a pointer this will result in shallow copying.3ClassName(const ClassName ¶meterName)If the class has at least one member variable that is a pointer it is necessary to:- replace the automatically generated copy construcor by implementing it- Use memberwise (copy‐by‐value) copying of member variables that are not pointers- Use deep copying for member variables that are pointers 练习，见 No.16 容器类知识点练习5、赋值号重载（Assignment operator）Each class has an assignment operator. Purpose is to create an identical copy of another object.Automatically generated if not defined. If a member variable is a pointer this will result in shallow copying.void operator=(const ClassName ¶meterName)If the class has at least one member variable that is a pointer it is necessary to:- replace the automatically generated assignmentoperator by implementing it- If the object is not assigned to itself Deallocate dynamically allocated memory Use memberwise (copy‐by‐value) copying of member variables that are not pointers Use deep copying for member variables that are pointers 练习，见 No.16 容器类知识点练习6、创建对象、new、基类指针利用默认构造函数创建对象（create an object using default constructor）用 new 创建对象，并赋值给类指针（create an object for the pointer）声明基类指针，并指向派生类对象（create an sub class object for the pointer which is a pointer of the base class type.） 练习，基于下面的类图，创建对象。

4Professional prof;Professional *profPtr = new Professional("E",2,"E");Runner * runnerPtr = NULL;runnerPtr = new Professional("D",1,"D");7、聚合关系 Relationship (composition/aggregation, has-a)A house ”has”– windows – aggregation, windows can exist without a house– door – aggregation, door can exist without a house– walls – composition, walls does not exist without a houseA Car “consists of/has” – a chassi (composition, 组成关系)– four wheels (aggregation, 聚合关系）A Car has 1 Chassi (composition) A Car has 4 Wheels (aggregation)The difference between aggregation and composition is that the Chassi will not exist without the Car but the wheels will.5 练习，聚合关系的判断7.1 A Student “is a” Person is an example of the relationship composition.7.2 A Car “has” four wheels is an example of the relationship composition.答案:7.1 FALSE, it is an example of the relationship inheritance. 7.2 FALSE, it is an example of the relationship aggregation, wheels can exist without a car.  练习，聚合关系的 C++实现方式在 Club.h 中添加动态数组的声明，该动态数组中保存 Member 对象或Member*类指针，两个属性 capacity 和 nrOfMember 分别表示数组容量和数组中当前保存的元素个数。

上图聚合关系的实现方案一（动态数组保存 Member 对象） ：Member * members; int capacity; int nrOfMembers;上图聚合关系的实现方案二（动态数组保存 Member*类指针）：Member * * members;int capacity; int nrOfMembers;8、继承关系 Relationship (Inheritance, is-a)base class（基类） , sub classes（派生类，子类）- An Employee is a Person- A Student is a Person- Things (properties and behavior) that are common for both Students and Employees is placed in the Person class- Things (properties and behavior) that are specific for Students are placed in the Student class6- Things (properties and behavior) that are specific for Employees are placed in the Employee class- Both Student and Employee inherits from Person9、继承的实现与函数调用继承的 C++实现方式：class SubClass : public BaseClass{…}- All members are inherited, though with different access- Public members in the base class are accessible in the subclass- Private members in base class are not accessible in the subclass 练习，类继承权限概念的判断9.1 Members in a sub class are accessible from the base class.9.2 Private members in a base class are accessible from a sub class.答案: 9.1 FALSE, a base class doesn’t know members of the sub class. 9.2 FALSE, -Private members in base class are not accessible in the subclass. 练习，类继承的实现与方法调用Employee 与 Person 之间的继承关系 C++是如何实现的？E。