如何在模板函数中根据类型构造不同的对象？

Question

这是我的问题，我有一个基础 class Base，两个派生的 classes ClassA 和 ClassB 形式 Base只接受两个参数，两个 classes ClassB0 和 ClassB1 派生自 ClassB 接受三个参数。我想根据传递给函数 create 的类型创建 class 实例，如果类型是从 ClassB 派生的，第一个参数将填充 10。编译器总是警告没有匹配的构造函数。

#include <iostream>
#include <type_traits>

class Base {};
class ClassA : public Base {
public:
  ClassA(int a, int b) : Base() {
    std::cout << "ClassA: " << a << " " << b << "\n\n";
  }
};

class ClassB : public Base {
public:
  ClassB(int a, int b, int c) : Base() {
    std::cout << "ClassB: " << a << " " << b << " " << c << "\n";
  }
};

class ClassB0 : public ClassB {
public:
  ClassB0(int a, int b, int c) : ClassB(a, b, c) { 
    std::cout << "ClassB0: " << a << " " << b << " " << c << "\n\n";
  }
};
class ClassB1 : public ClassB {
public:
  ClassB1(int a, int b, int c) : ClassB(a, b, c) {
    std::cout << "ClassB1: " << a << " " << b << " " << c << "\n\n";
  }
};

template <typename T, typename ...Args>
T* create(Args&&... args) {
  T* comp = nullptr;

  if (std::is_base_of<ClassB, T>::value) {
    std::cout << "True ";
    comp = new T(10, std::forward<Args>(args)...);
  } else {
    std::cout << "False ";
    comp = new T(std::forward<Args>(args)...);
  }

  return comp;
}

int main() {
  create<ClassA>(1, 2);
  create<ClassB0>(2, 3);
  create<ClassB1>(2, 3);
}

Answer 1

问题是 if 和 else 的分支都需要在编译时求值，尽管其中一个会在运行时求值。

您可以应用 constexpr if statement (C++17 起)，其条件必须在编译时已知，并且 if 或 else 分支将是被丢弃，不会在编译时再次求值。

If the value is true, then statement-false is discarded (if present), otherwise, statement-true is discarded.

template <typename T, typename ...Args>
T* create(Args&&... args) {
  T* comp = nullptr;

  if constexpr (std::is_base_of<ClassB, T>::value) {
  // ^^^^^^^^^
    std::cout << "True ";
    comp = new T(10, std::forward<Args>(args)...);
  } else {
    std::cout << "False ";
    comp = new T(std::forward<Args>(args)...);
  }

  return comp;
}

LIVE

在 C++17 之前，您可以使用 SFINAE 应用重载。例如

template <typename T, typename ...Args>
typename std::enable_if<std::is_base_of<ClassB, T>::value, T*>::type
create(Args&&... args) {
  T* comp = nullptr;
  std::cout << "True ";
  comp = new T(10, std::forward<Args>(args)...);
  return comp;
}
template <typename T, typename ...Args>
typename std::enable_if<!std::is_base_of<ClassB, T>::value, T*>::type
create(Args&&... args) {
  T* comp = nullptr;
  std::cout << "False ";
  comp = new T(std::forward<Args>(args)...);
  return comp;
}

LIVE

Answer 2

也许我弄错了，听起来你想实现工厂模式。您有一个 class 层次结构，并希望基于键创建派生的 class。

您遇到的困难是构造函数具有不同数量的参数。

在下面的代码中，我使用可变模板和 std::any.

完全解决了你的问题

请参阅下面的完整工作示例：

#include <iostream>
#include <map>
#include <utility>
#include <any>


// Some demo classes ----------------------------------------------------------------------------------
struct Base {
    Base(int d) : data(d) {};
    virtual ~Base() { std::cout << "Destructor Base\n"; }
    virtual void print() { std::cout << "Print Base\n"; }
    int data{};
};
struct Child1 : public Base {
    Child1(int d, std::string s) : Base(d) { std::cout << "Constructor Child1 " << d << " " << s << "\n"; }
    virtual ~Child1() { std::cout << "Destructor Child1\n"; }
    virtual void print() { std::cout << "Print Child1: " << data << "\n"; }
};
struct Child2 : public Base {
    Child2(int d, char c, long l) : Base(d) { std::cout << "Constructor Child2 " << d << " " << c << " " << l << "\n"; }
    virtual ~Child2() { std::cout << "Destructor Child2\n"; }
    virtual void print() { std::cout << "Print Child2: " << data << "\n"; }
};
struct Child3 : public Base {
    Child3(int d, long l, char c, std::string s) : Base(d) { std::cout << "Constructor Child3 " << d << " " << l << " " << c << " " << s << "\n"; }
    virtual ~Child3() { std::cout << "Destructor Child3\n"; }
    virtual void print() { std::cout << "Print Child3: " << data << "\n"; }
};



using UPTRB = std::unique_ptr<Base>;


template <class Child, typename ...Args>
UPTRB createClass(Args...args) { return std::make_unique<Child>(args...); }

// The Factory ----------------------------------------------------------------------------------------
template <class Key, class Object>
class Factory
{
    std::map<Key, std::any> selector;
public:
    Factory() : selector() {}
    Factory(std::initializer_list<std::pair<const Key, std::any>> il) : selector(il) {}

    template<typename Function>
    void add(Key key, Function&& someFunction) { selector[key] = std::any(someFunction); };

    template <typename ... Args>
    Object create(Key key, Args ... args) {
        if (selector.find(key) != selector.end()) {
            return std::any_cast<std::add_pointer_t<Object(Args ...)>>(selector[key])(args...);
        }
        else return nullptr;
    }
};

int main()
{
    Factory<int, UPTRB> factory{
        {1, createClass<Child1, int, std::string>},
        {2, createClass<Child2, int, char, long>}
    };
    factory.add(3, createClass<Child3, int, long, char, std::string>);


    // Some test values
    std::string s1(" Hello1 "); std::string s3(" Hello3 ");
    int i = 1;  const int ci = 1;   int& ri = i;    const int& cri = i;   int&& rri = 1;

    UPTRB b1 = factory.create(1, 1, s1);
    UPTRB b2 = factory.create(2, 2, '2', 2L);
    UPTRB b3 = factory.create(3, 3, 3L, '3', s3);

    b1->print();
    b2->print();
    b3->print();
    b1 = factory.create(2, 4, '4', 4L);
    b1->print();
    return 0;
}

如何在模板函数中根据类型构造不同的对象？

How to construct different object according to their types in a template function?

c++

templates

variadic-templates

parameter-pack