【发布时间】：2017-01-05 09:50:31
【问题描述】：
在学习了这里的一些优秀教程之后，我正在尝试创建一个简单的状态模式：http://gameprogrammingpatterns.com/state.html
当前教程已经完成一半，我正在尝试复制每个状态的静态实例，方法是将它们包含在基类中。但是，当涉及到切换状态时，g++ 会抛出此错误。
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，我了解该错误涉及指针的转换，但我真的很想知道如何解决它。当我在关注这些人的代码时，我有点希望它能够工作，但是因为他正在改变它并试图加强最佳实践，所以我没有他完整的源代码可以遵循。但是，我觉得在完成本教程的其余部分之前，了解此阶段的代码对我来说很重要。
以下是我创建的代码，试图复制他的状态系统：
#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没有编译？
编辑：
因此，我非常感谢之前的辅导，因此我修改了代码。我首先将它们放在单独的文件中，但是对于如此少量的测试代码来说，这比它的价值要麻烦得多。我只是重写了一个声明类的头文件，然后在 .cpp 文件中定义它们。
这是.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;
}
我仍然无法让它工作。以前的错误已经消失，但我正在努力访问“介绍”的静态实例并在基类中播放。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
我以为我搞砸了！好沮丧！
我应该补充一点，RustyX 提供的答案确实可以编译，但是我必须将“播放”和“介绍”的实例移到类定义之外，然后我不能再将它们设置为静态，我相信这很重要，因为我只需要每个实例一个，我希望它们尽早初始化。
【问题讨论】：
-
你可能对我的STTCL 框架感兴趣。
标签： c++ class virtual state-pattern