class Car {
class BaseState {
explicit BaseState(Car* vehicle) : mVehicle(vehicle) {}
virtual void run() = 0;
Car* mVehicle;
}
class State1 : public BaseState {
explicit State1(Car* vehicle) : BaseState(vehicle) {}
virtual void run() {
// use data of Car
...
doSomething();
}
virtual void doSomething() {
}
}
class State2 : public BaseState {
}
...
}
class Convertible: public Car {
class State1 : public Car::State1 {
explicit State1(Convertible* vehicle) : Car::State1(vehicle) {}
virtual void doSomething() {
static_cast<Convertible*>(mVehicle)->foldTop();
}
}
class State2 : public Car::State2 {
}
...
void foldTop() {}
}
所有状态都从BaseState派生，因此它们具有成员变量mVehicle来访问外部类变量。
但是，在每个派生类中，在每个状态的所有函数中，需要static_cast来访问派生类成员变量和函数。
有更好的解决方案吗？
================================================== =====================
是的。我尝试了如下模板，但由于诸如以下错误而无法编译
“car.h:在成员函数“虚拟无效的Car::State1::run()”中:
car.h:18:12:错误:未在此范围内声明“mVehicle”
”。
// car.h
#include <iostream>
template <class T>
class Car {
public:
class BaseState {
public:
explicit BaseState(T* vehicle) : mVehicle(vehicle) {}
protected:
T* mVehicle;
};
class State1 : public BaseState {
public:
explicit State1(T* vehicle) : BaseState(vehicle) {}
virtual void run() {
mVehicle->x = 1;
mVehicle->y = 2;
mVehicle->doSomething1();
mVehicle->doSomething2();
processEvent();
}
virtual void processEvent() {
if (mVehicle->val > 2) {
std::cout << "too large" << std::endl;
}
}
};
class State2 : public BaseState {
public:
explicit State2(T* vehicle) : BaseState(vehicle) {}
virtual void run() {
mVehicle->x = 10;
mVehicle->y = 20;
processEvent();
}
virtual void processEvent() {
if (mVehicle->val > 20) {
std::cout << "too large" << std::endl;
}
}
};
virtual void doSomething1() {
val += x * y;
}
virtual void doSomething2() {
val += x + y;
}
protected:
int x;
int y;
int val;
};
// convertible.h
#include "car.h"
#include <iostream>
class Convertible : public Car<Convertible> {
protected:
class State1 : public Car<Convertible>::State1 {
explicit State1(Convertible* vehicle) : Car<Convertible>::State1(vehicle) {}
// want to override functions in base class states
virtual void processEvent() {
if (mVehicle->val > 10) {
std::cout << "too large" << std::endl;
mVehicle->val = 10;
}
}
};
// want to override some base class functions
// and access some special variables
// want to inherit other functions
virtual void doSomething2() {
z = 10;
val += x + y + z;
}
protected:
int z;
};
如果我使用
State1(Car* vehicle)而不是State1(T* vehicle)，则存在其他转换错误。我究竟做错了什么？如果程序能够确定应该执行
Convertible::State1::processEvent()，为什么它不能自动将mVehicle从Car*转换为Convertible*？当推导mVehicle时，Convertible显然指向Convertible::State1::processEvent()对象。如果有自动转换，则不需要模板。 最佳答案
此实现不使用强制类型转换，重复的指针，虚拟 getter 或CRTP。它具有三个并行层次结构:
所以我们有
Car Car::AbstractState Car::State<C>
| | |
+--- Convertible +--- Convertible::AbstractState +--- Convertible::State<C>
| | | | | |
| +--- Racer | +--- Racer::AbstractState | +--- Racer::State<C>
+--- Hybrid +--- Hybrid::AbstractState +--- Hybrid::State<C>
每个具体状态都源自并实现相应的抽象状态。如果我们有一个指向
Car*的Convertible，并查询其状态，我们将得到一个Car::AbstractState*，它指向最终状态为Convertible::State<Convertible>的具体状态对象。但是，汽车层次结构的用户不知道也不关心模板机制。编码:
#include <iostream>
using namespace std;
struct Trace
{
Trace(const char* s) : s (s)
{
cout << s << " start\n";
}
~Trace()
{
cout << s << " end\n";
}
const char* s;
};
struct Car {
struct AbstractState
{
virtual void run() = 0;
};
template <typename C>
struct State : virtual AbstractState
{
explicit State(C* vehicle) : mVehicle(vehicle) {}
virtual void run()
{
Trace("Car::State::run");
doSomething();
};
virtual void doSomething()
{
Trace("Car::State::doSomething");
}
C* mVehicle;
};
virtual AbstractState* getState() { return new State<Car>(this); }
};
struct Convertible : Car {
struct AbstractState : virtual Car::AbstractState
{
virtual void runBetter() = 0;
};
template <typename C>
struct State : Car::State<C>, virtual AbstractState
{
using Car::State<C>::mVehicle;
explicit State(C* vehicle) : Car::State<C>(vehicle) {}
void doSomething()
{
Trace("Convertible::State::doSomething");
Car::State<C>::doSomething();
mVehicle->foldTop();
}
void runBetter()
{
Trace("Convertible::State::runBetter");
run();
doSomethingElse();
};
virtual void doSomethingElse()
{
Trace("Convertible::State::doSomethingElse");
}
};
void foldTop()
{
Trace("Convertible::foldTop");
}
Convertible::AbstractState* getState() { return new State<Convertible>(this); }
};
int main ()
{
Car car;
Convertible convertible;
Car& car2(convertible);
cout << "runing car\n";
Car::AbstractState* carstate = car.getState();
carstate->run();
cout << "runing convertible\n";
Convertible::AbstractState* convertiblestate = convertible.getState();
convertiblestate->run();
cout << "runing car2\n";
Car::AbstractState* carstate2 = car2.getState();
carstate2->run();
}
关于c++ - 访问虚函数中的派生类成员变量，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/33010677/