抽象类构造函数问题

C/C++ code

enum Size{Mini, Medium, Full};class CVehicle{public:    CVehicle(Size s = Medium, int n = 0) : size(s), nSpotsNeeded(n) {}protected:    Size size;    int    nSpotsNeeded;};class CBus : public CVehicle{public:    CBus() : CVehicle(Full, 5) {}};

请问怎么在没其他虚函数的情况下把基类CVehicle声明为抽象类。。构造函数显然不能成为虚函数，还有其他办法么？是否把析构函数变纯虚？

这种没多态动作的继承是不是不好?

[解决办法]
可以。
不过注意对于析构函数，即便是纯虚函数，也需要定义，否则没法odr-use，用了链接错误。（普通纯虚成员函数的定义倒是可有可无。）

[解决办法]

C/C++ code

#include <iostream>using namespace std;enum Size{Mini, Medium, Full};class CVehicle{public:    CVehicle(Size s = Medium, int n = 0) : size(s), nSpotsNeeded(n) {}    virtual~CVehicle()=0;//完全可以把析构函数设置为纯虚函数，编译能够通过的。protected:    Size size;    int    nSpotsNeeded;};class CBus : public CVehicle{public:    CBus() : CVehicle(Full, 5) {}};void main(){}
[解决办法]
探讨

C/C++ code

enum Size{Mini, Medium, Full};
class CVehicle{
public:
    CVehicle(Size s = Medium, int n = 0) : size(s), nSpotsNeeded(n) {}
protected:
    Size size;
    int    nSpotsNeeded;
};

class ……