提问人：Fzza 提问时间：10/26/2023 更新时间：10/26/2023 访问量：46
创建动态推导类型的智能指针？
Create a smart pointer of a dynamically deduced type?
问：
是否可以在不使用 if 序列的情况下创建动态推断类型的指针？
我应该使用什么作为地图值类型？
class Vehicle {
public:
virtual void run() =0;
};
// Derived class
class Mustang: public Vehicle {
public:
string model = "Mustang";
virtual void run() {std::cout<<model<<std::endl;};
};
// Derived class
class Ford: public Vehicle {
public:
string model = "Ford";
virtual void run() {std::cout<<model<<std::endl;};
};
std::unordered_map<std::string, std::type_index> m {{"Mustang",typeid(Mustang)},{"Ford",typeid(Ford)} };
int main()
{
std::unique_ptr<Vehicle> v;
std::string s{"Mustang"};
v=std::make_unique<m.at(s)>;
v->run(); //what i want to achieve is print "Mustang"
return 0;
}
答：
2赞
463035818_is_not_an_ai
10/26/2023
#1
您可以在映射中存储返回 a 的可调用对象：std::unique_ptr<Vehicle>
#include <string>
#include <memory>
#include <iostream>
#include <unordered_map>
#include <functional>
class Vehicle {
public:
virtual void run() =0;
virtual ~Vehicle() {}
};
// Derived class
class Mustang: public Vehicle {
public:
std::string model = "Mustang";
virtual void run() {std::cout<<model<<std::endl;};
};
// Derived class
class Ford: public Vehicle {
public:
std::string model = "Ford";
virtual void run() {std::cout<<model<<std::endl;};
};
std::unordered_map<std::string, std::function<std::unique_ptr<Vehicle>()>> m {
{"Mustang",[]()->std::unique_ptr<Vehicle> { return std::make_unique<Mustang>();}},
{"Ford",[]() ->std::unique_ptr<Vehicle> { return std::make_unique<Ford>();}}
};
int main()
{
std::unique_ptr<Vehicle> v;
std::string s{"Mustang"};
v=m.at(s)();
//^^---------- note the () to call the function
// that returns the std::unique_ptr
v->run(); // prints Mustang
return 0;
}
这：
[]()->std::unique_ptr<Vehicle> { return std::make_unique<Mustang>();}
相当冗长，因此您可能需要编写一个辅助函数：
template <typename T>
std::unique_ptr<Vehicle> create_vehicle() {
return std::make_unique<T>();
}
PS：您缺少虚拟析构函数。Vehicle
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