Question

我是C ++的新手。我用SFML和Box2D开发了一个突破克隆，编译时我遇到了这个错误。错误详情：

c:\program files (x86)\visual studio express 2013\vc\include\xutility(2420): error C2280: 'Tile &Tile::operator =(const Tile &)' : attempting to reference a deleted function
c:\users\harry\documents\visual studio 2013\projects\cpp-breakout\cpp-breakout\entities\tile.hpp(27) : compiler has generated 'Tile::operator =' here

当我尝试从std :: vector＆lt;中删除Tile对象时出现错误。瓦片GT; with vector.erase。

Tile.h

class Tile : public Entity {
public:
    Tile(b2World* world, float posX, float posY);
    void draw(sf::RenderWindow& window);

    const float PTM_RATIO = 32.f;
    const int HALF_WIDTH = 32;
    const int HALF_HEIGHT = 16;
    int armor;
    bool flaggedToErase = false;

    b2Body* tileBody;
    sf::Sprite sprite;
};

Tile.cpp

Tile::Tile(b2World* world, float posX, float posY)
{
    // Define a body.
    b2BodyDef tileBodyDef;
    tileBodyDef.type = b2_staticBody;
    tileBodyDef.position.Set((posX + HALF_WIDTH) / PTM_RATIO, (posY + HALF_HEIGHT) / PTM_RATIO);

    // Use the body definition to create the actual body instance.
    tileBody = world->CreateBody(&tileBodyDef);
    tileBody->SetUserData(this);
    // Define shape.
    b2PolygonShape tileShape;
    tileShape.SetAsBox(HALF_WIDTH / PTM_RATIO, HALF_HEIGHT / PTM_RATIO);

    // Define fixture.
    b2FixtureDef tileFixtureDef;
    tileFixtureDef.shape = &tileShape;
    tileFixtureDef.density = 10.0f;
    tileFixtureDef.restitution = 0.1f;
    tileFixtureDef.friction = 0.0f;
    //tileFixtureDef.isSensor = true;

    bUserData* bud = new bUserData;
    bud->entityType = TILE;
    tileFixtureDef.userData = bud;

    // Create fixture.
    tileBody->CreateFixture(&tileFixtureDef);
}

我将Tiles推送到

中的向量

Level.cpp

std::vector<Tile> solidTiles;

void Level::loadLevel(b2World* world) {
    // Loop through the map and set the tile position and sprites to the tiles.
    for(unsigned int i = 0; i < map.size(); i++) {
        for(unsigned int j = 0; j < map[i].size(); j++) {
            // Set sprites to the tiles in every grid cell which is not -1,-1.
            if(map[i][j].x != -1 && map[i][j].y != -1) {
                Tile tempTile(world, j * TILE_WIDTH, i * TILE_HEIGHT);
                tempTile.sprite = tiles;
                tempTile.sprite.setOrigin(sf::Vector2f(HALF_WIDTH, HALF_HEIGHT));
                tempTile.sprite.setTextureRect(sf::IntRect(map[i][j].x * TILE_WIDTH, map[i][j].y * TILE_HEIGHT, TILE_WIDTH, TILE_HEIGHT));
                tempTile.sprite.setPosition(sf::Vector2f(tempTile.tileBody->GetPosition().x * PTM_RATIO, tempTile.tileBody->GetPosition().y * PTM_RATIO));
                solidTiles.push_back(tempTile);
            }
        }
    }
}

我尝试擦除

中的图块

PlayState.cpp

void PlayState::removeSprites() {
    for(unsigned int i = 0; i < level1.solidTiles.size(); i++) {
        if(level1.solidTiles[i].flaggedToErase == true) {
            level1.solidTiles.erase(level1.solidTiles.begin() + i);
        }
    }
}

问题是否与移动构造函数/赋值有关？

Answer 1

由于Tile类具有 const数据成员，因此编译器复制assignmnet运算符隐式定义它被定义为已删除。

您必须自己定义复制赋值运算符显式。

Answer 2

错误明确提到Tile &Tile::operator =(const Tile &)，所以它是关于赋值运算符，而不是移动赋值运算符。您需要为Tile实现它。你也缺少Tile的拷贝构造函数。

引自here：

隐式声明或默认的复制赋值运算符   类T被定义为删除以下任何一种情况：


T具有const
的非静态数据成员
T具有引用类型的非静态数据成员。

T具有无法复制分配的非静态数据成员（已删除，无法访问或模糊的复制赋值运算符）

T具有无法复制分配的直接或虚拟基类（已删除，不可访问或模糊的移动赋值运算符）

T具有用户声明的移动构造函数

T具有用户声明的移动赋值运算符

错误C2280：尝试引用已删除的函数（尝试调用vector.erase）

2 个答案: