7
我试图挑起一个简单的状态模式，下面的一些优秀的教程这里经过：http://gameprogrammingpatterns.com/state.html状态模式C++
我通过这个电流教程半路上，我试图复制的静态实例每个状态，通过将它们包含在基类中。但是，当谈到切换状态时，g ++会抛出这个错误。
state_test.cpp: In member function ‘virtual void Introduction::handleinput(Game&, int)’:
state_test.cpp:55:16: error: cannot convert ‘Playing*’ to ‘GameState*’ in assignment
game.state_ = &GameState::play;
^
现在，我明白错误涉及指针的转换，但我真的很难看到如何解决它。正如我遵循这个家伙的代码，我期望它能够工作，但是因为他正在改变它并试图强化最佳实践，所以我没有完整的源代码可以遵循。但是，在我继续阅读本教程的其余部分之前，我认为在此阶段理解代码非常重要。
以下是我创建的代码试图复制他的国家制度：
#include <iostream>
class Game;
class Introduction;
class Playing;
class GameState
{
public:
static Introduction intro;
static Playing play;
virtual ~GameState() {std::cout << "an undefined GameState has been destroyed" << std::endl;}
virtual void handleinput(Game& game, int arbitary) {}
virtual void update(Game& game) {}
};
class Game
{
public:
Game()
{}
~Game()
{}
virtual void handleinput(int arbitary)
{
state_->handleinput(*this, arbitary);
}
virtual void update()
{
state_->update(*this);
}
//private:
GameState* state_;
};
class Introduction : public GameState
{
public:
Introduction()
{
std::cout << "constructed Introduction state" << std::endl;
}
virtual void handleinput(Game& game, int arbitary)
{
if (arbitary == 1)
game.state_ = &GameState::play;
}
virtual void update(Game& game) {}
};
class Playing : public GameState
{
public:
Playing() {std::cout << "constructed Playing state" << std::endl;}
virtual void handleinput(Game& game, int arbitary)
{
if (arbitary == 0)
game.state_ = &GameState::intro;
}
virtual void update(Game& game) {}
};
int main(int argc, char const *argv[])
{
Game thisgame;
return 0;
}
任何想法，为什么我的实现不是编制？
编辑：
所以响应早前辅导，对此我很感激，我修改了代码。我首先将它们全部放在单独的文件中，但这比测试代码量小的代价更麻烦。我简单地重写了一个声明类的头文件，然后在.cpp文件中定义它们。
这里是.h文件：
class Introduction;
class Playing;
class Game;
class GameState;
class GameState
{
public:
static Introduction intro;
static Playing play;
virtual ~GameState();
virtual void handleinput(Game& game, int arbitary);
virtual void update(Game& game);
};
class Introduction : public GameState
{
public:
Introduction();
virtual void handleinput(Game& game, int arbitary);
virtual void update(Game& game);
};
class Playing : public GameState
{
public:
Playing();
virtual void handleinput(Game& game, int arbitary);
virtual void update(Game& game);
};
class Game
{
public:
Game();
~Game();
virtual void handleinput(int arbitary);
virtual void update();
GameState* state_;
};
这里是.cpp文件：
#include <iostream>
#include "state.h"
GameState::~GameState()
{std::cout << "Exiting Game State Instance" << std::endl;}
void GameState::handleinput(Game& game, int arbitary)
{}
void GameState::update(Game& game)
{}
Game::Game()
{}
Game::~Game()
{}
void Game::handleinput(int arbitary)
{
state_->handleinput(*this, arbitary);
}
void Game::update()
{
state_->update(*this);
}
Introduction::Introduction()
{
std::cout << "constructed Introduction state" << std::endl;
}
void Introduction::handleinput(Game& game, int arbitary)
{
if (arbitary == 1)
game.state_ = &GameState::play;
}
void Introduction::update(Game& game) {}
Playing::Playing()
{
std::cout << "constructed Playing state" << std::endl;
}
void Playing::handleinput(Game& game, int arbitary)
{
if (arbitary == 0)
game.state_ = &GameState::intro;
}
void Playing::update(Game& game) {}
int main(int argc, char const *argv[])
{
Game mygame;
return 0;
}
而且我仍然无法得到它的工作。之前的错误已经消失，但我正在努力访问“引入”的静态实例并在基类内部进行游戏。抛出的错误是：
/tmp/ccH87ioX.o: In function `Introduction::handleinput(Game&, int)':
state_test.cpp:(.text+0x1a9): undefined reference to `GameState::play'
/tmp/ccH87ioX.o: In function `Playing::handleinput(Game&, int)':
state_test.cpp:(.text+0x23f): undefined reference to `GameState::intro'
collect2: error: ld returned 1 exit status
我以为我有它被怀疑！太沮丧了！
我应该补充一点，RustyX提供的答案确实会编译，但是我必须将“播放”和“引入”的实例移到类定义之外，然后我不能再将它们设置为静态的，我相信这很重要，因为我只需要每个实例的一个实例，我希望它们尽早初始化。
您可能对我的[STTCL]（https://github.com/makulik/sttcl）框架感兴趣。 –