我有一个叫做游戏的类,这是它的 operator=: 的代码
Game& Game::operator=(const Game &other){
if(this==&other){
return *this;
}else{
for(unsigned i=0;i<other.players.size();i=i+1){
Player* copy;
int str= other.players.at(i)->getStr();
if(str==1)
copy = new PlayerType1(dynamic_cast<const PlayerType1&>(*other.players.at(i)));
if(str==2)
copy = new PlayerType2(dynamic_cast<const PlayerType2&>(*other.players.at(i)));
if(str==3)
copy = new PlayerType3(dynamic_cast<const PlayerType3&>(*other.players.at(i)));
if(str==4)
copy = new PlayerType4(dynamic_cast<const PlayerType4&>(*other.players.at(i)));
players.push_back(copy);
}
winners = other.winners;
state = vector<string>(other.state);
deck = Deck(other.deck);
verbal = other.verbal;
highestNum = other.highestNum;
turnNum = other.turnNum;
currPlayer = other.currPlayer;
lastAsker = other.lastAsker;
lastAskee = other.lastAskee;
lastAskedCard = other.lastAskedCard;
return *this;
}
我尝试在这里调用它:
char* cf= "../src/config1.txt";
Game* game = new Game(cf);
game->init();
Game game2=*game;
game->play();
game2.printState();
在这种情况下,我的 operator= 将不会被使用。 但是如果 game2 已经初始化,例如这里:
Game* game = new Game(cf);
game->init();
Game game2=*(new Game());
game2=*game;
game->play();
game2.printState();
知道可能是什么问题吗?
最佳答案
在第一种情况下,copy elision避免调用赋值运算符,支持复制构造函数。
在第二种情况下,对象已经构建,因此必须调用赋值运算符。
结论是您必须实现所有 3 个运算符:析构函数、赋值和复制(也称为 rule of three),否则您可能会出现不可预测的行为,具体取决于编译器。
避免以不一致的方式实现它们的最佳方法是使用 copy & swap idiom
关于c++ - Operator= 除非对象已经初始化,否则重载不起作用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40821437/