我想重新分配一个具有自引用的类,名为CarJoker
。
“重新分配”在这里意味着=更改对象的地址。
此技术对于使CarJoker
的每个实例都位于可调整大小的连续数组(例如,数组)中是必需的。池。
我考虑使用std::move
,但是它不能以我希望的方式移动CarJoker::wheels
。
(MCVE)
#include <vector>
#include <iostream>
#include <string>
struct Wheel{
void setBlade(){}
void setTwinkle(){}
};
struct CarJoker{
Wheel wa;
Wheel wb;
Wheel wc;
std::vector<Wheel*> wheels;
float hp=5;
CarJoker(){
wheels.push_back(&wa);
wheels.push_back(&wb);
wheels.push_back(&wc);
}
void wow(){
//v want to apply something to every "wheel"
for(auto wheel:wheels){
wheel->setBlade();
}
//v want to apply something to some certain "wheel"
wa.setTwinkle();
}
};
int main(){
CarJoker car1;
CarJoker car2=std::move(car1);
std::cout<<"want to 1 : "<<(car2.wheels[0]== &car2.wa)<<std::endl;
}
使用
std::move
时,car2.wheels[0]
指向&car1.wa
而不是我希望的&car2.wa
。我知道原因,但这不是我的目标,而且我不知道解决问题的优雅方法。
我可怜的解决方法
这是一种优雅的方式(MCVE):
struct CarJoker{
Wheel wa;
Wheel wb;
Wheel wc;
std::vector<Wheel*> wheels;
float hp=5;
CarJoker(){
reIni(); //: CHANGE (call a new function)
}
void reIni(){ //: CHANGE (create a new function)
wheels.clear();
wheels.push_back(&wa);
wheels.push_back(&wb);
wheels.push_back(&wc);
}
void wow(){
//v want to apply something to every "wheel"
for(auto wheel:wheels){
wheel->setBlade();
}
//v want to apply something to some certain "wheel"
wa.setTwinkle();
}
};
int main(){
CarJoker car1;
CarJoker car2=std::move(car1);
car2.reIni(); //: CHANGE (call a new function)
std::cout<<"want to 1 : "<<(car2.wheels[0]== &car2.wa)<<std::endl;
}
坏处:-
1.很脏。
2.我必须为池中存在此类症状的每个类创建一个特殊名称函数(
reIni()
)。我的池也必须识别该功能(例如,使用模板或虚拟功能进行注册)。我可怜的解决方法2
struct CarJoker{
Wheel wa;
Wheel wb;
Wheel wc;
std::vector<Wheel*> getWheels(){ //use this instead of "wheels"
std::vector<Wheel*> re;
re.push_back(&wa);
re.push_back(&wb);
re.push_back(&wc);
return re;
}
....
}
我会工作,但我觉得这样的解决方法很疯狂。
限制增加了编码人员的陷阱。
如果
wheels
碰巧需要缓存计算量大的结果,那么现在经常调用getWheels()
会很昂贵。 最佳答案
您不需要reIni()
方法。您需要添加:
复制构造函数和移动构造函数,它们都以与默认构造函数相同的方式初始化wheels
成员。
不复制/移动wheels
成员的副本分配运算符和移动分配运算符。
尝试这个:
struct CarJoker{
Wheel wa;
Wheel wb;
Wheel wc;
std::vector<Wheel*> wheels;
float hp = 5;
CarJoker(){
wheels.push_back(&wa);
wheels.push_back(&wb);
wheels.push_back(&wc);
}
CarJoker(const CarJoker &src) :
CarJoker(),
wa(src.wa),
wb(src.wb),
wc(src.wc),
//wheels(src.wheels),
hp(src.hp){
}
CarJoker(CarJoker &&src) :
CarJoker(),
wa(std::move(src.wa)),
wb(std::move(src.wb)),
wc(std::move(src.wc)),
//wheels(std::move(src.wheels)),
hp(src.hp){
}
// copy assignment and move assignment can be
// handled with a single implementation that
// lets the compiler choose between the copy
// constructor and move constructor as needed...
CarJoker& operator=(CarJoker rhs){
wa = std::move(rhs.wa);
wb = std::move(rhs.wb);
wc = std::move(rhs.wc);
wheels = std::move(rhs.wheels);
hp = rhs.hp;
return *this;
}
...
};
话虽如此,您真的不应该使用自引用字段作为开始。单个
std::array<Wheel, 3>
字段比3个Wheel
字段和std::vector<Wheel*>
字段有意义,避免了整个问题。struct CarJoker{
std::array<Wheel, 3> wheels;
float hp = 5;
CarJoker() = default;
CarJoker(const CarJoker&) = default;
CarJoker(CarJoker&&) = default;
CarJoker& operator=(const CarJoker&) = default;
CarJoker& operator=(CarJoker&&) = default;
Wheel& wa() { return wheels[0]; }
const Wheel& wa() const { return wheels[0]; }
Wheel& wb() { return wheels[1]; }
const Wheel& wb() const { return wheels[1]; }
Wheel& wc() { return wheels[2]; }
const Wheel& wc() const { return wheels[2]; }
void wow(){
//v want to apply something to every "wheel"
for(auto &wheel : wheels){
wheel.setBlade();
}
//v want to apply something to some certain "wheel"
wa().setTwinkle();
}
};