c++ public 成员仅在构造函数具有参数时才可访问

Question

所以我创建了这个 class，其中包含一些 public 成员，它们是布尔值，但是，当我实际尝试使用默认构造函数（无参数）创建对象时，我无法访问这些成员。但是，如果我用参数 (unsigned int) 声明构造函数，我就可以访问成员。有人知道为什么吗？

以下为class：

class MoveManagement {

private:
    Player * currentPlayer;
    sf::View* currentView;
    int lambda_x = 0;
    int lambda_y = 0;

public:
    bool m_up;
    bool m_down;
    bool m_left;
    bool m_right;

    MoveManagement() {
        m_up = false;
        m_down = false;
        m_left = false;
        m_right = false;
    }

    void getNextView(Player* player_, sf::View* view_) {
        currentPlayer = player_;
        currentView = view_;

        if (m_up) {
            --lambda_y;
        }

        if (m_down) {
            ++lambda_y;
        }

        if (m_left) {
            --lambda_x;
        }

        if (m_right) {
            ++lambda_x;
        }

        currentPlayer->playerCharacter->m_position.x = currentPlayer->playerCharacter->m_position.x + lambda_x;
        currentPlayer->playerCharacter->m_position.y = currentPlayer->playerCharacter->m_position.y + lambda_y;

        currentView->move(lambda_x, lambda_y);

        lambda_x = 0;
        lambda_y = 0;
    }
};

我这样创建一个新对象：

MoveManagement move();

如果我尝试访问任何成员，我会收到一条错误消息，提示 "expression must have class type"。

谢谢。

Answer 1

MoveManagement move(); 声明了一个函数。请改用 MoveManagement move;。

c++ public 成员仅在构造函数具有参数时才可访问

c++ public members only accessible if constructor has a parameter

c++

most-vexing-parse