Question

我有3个类1）GameStateManager 2）MenuState（从另一个名为GameState的类继承）3）GameState

我想将此指针作为参数传递给MenuState的构造函数。

我的目标是在MenuState中获取指向GameStateManager对象的指针以供将来使用。

我收到错误：

error C2061: syntax error: identifier 'GameStateManager'
error C2664: 'MenuState::MenuState(const MenuState &)': cannot convert argument 1 from 'GameStateManager' to 'const MenuState &'
note: Reason: cannot convert from 'GameStateManager' to 'const MenuState'

GameStateManager.h

#pragma once

#include <vector>
#include <SFML/Graphics.hpp>
#include "MenuState.h"

class GameStateManager {
public:
    static const int MENUSTATE = 0;
    static const int FIRSTLEVELSTATE = 1;

    GameStateManager();

    ~GameStateManager();
private:
    std::vector<GameState*> States;
    int currentState;
};

GameStateManager.cpp

GameStateManager::GameStateManager() {
    this->currentState = MENUSTATE;
    this->States.push_back(new MenuState(*this)); // Error Line. I Think!
}

MenuState.h

#pragma once
#include "GameState.h"
#include "GameStateManager.h"
class MenuState: public GameState{
public:
    MenuState(GameStateManager& gsm);
    ~MenuState();
};

MenuState.cpp

MenuState::MenuState(GameStateManager& gsm){ // Error Line. I Think!
}

GameState.h

#pragma once
#include <SFML/Graphics.hpp>
class GameState {
//Virtual methods are here in this code which are not important for this question
public:
    GameState() {}
    ~GameState() {}
};

Main.cpp的

#include <SFML/Graphics.hpp>
#include "GameStateManager.h"

int main(int argc, char** argv) {
    GameStateManager gsm;
}

Answer 1

由于GameStateManager.h和MenuState.h

中的标题包含冲突

如果声明为指针或引用，则使用前向声明：

MenuState::MenuState(class GameStateManager& gsm);
std::vector<class MenuState*> States;

然后，将行#include "GameStateManager.h"放入MenuState.cpp并将行#include "MenuState.h"放入GameStateManager.h将减少错误。

错误将此指针作为参数传递给类构造函数

1 个答案: