是否可以编写C ++基类成员函数来实例化从中调用的任何派生类？

Question

有没有办法只在基类中实现一个函数，该函数将返回从中调用它的任何派生类的实例？

Answer 1

template <class TDerived> class Base
{
    TDerived retInstance()
    {
        return TDerived();
    }
};

class Derived : Base<Derived>
{
    //class definition here
};

Answer 2

如上所述，

1. CRTP（奇怪的重复模板模式）

2. 或Cloneable Pattern。

---

1。 CRTP

查看 Liv on Coliru

template <typename Derived>
struct BaseImpl
{
    // normal stuff
    int foo() const { return 42; }

    // CRTP stuff
    Derived make_new() const 
    {
        return Derived("test 123");
    }
};

#include <string>

struct MyStruct : BaseImpl<MyStruct>
{
    std::string value;
    MyStruct(std::string const& value) : value(value) {}
};

#include <iostream>

int main()
{
    MyStruct a("first");
    MyStruct b = a.make_new();

    std::cout << a.value << "\n"
              << b.value << "\n";
}

打印

first
test 123

---

2。可克隆模式：

同时查看 Live on Coliru

struct ICloneable
{
    virtual const char* whoami() const = 0;
    virtual ICloneable* clone() const = 0;

    virtual ~ICloneable() throw() {}
};

#include <string>

struct A : ICloneable
{
    virtual const char* whoami() const { return "struct A"; }
    virtual ICloneable* clone() const { return new A; }
};

struct B : ICloneable
{
    virtual const char* whoami() const { return "struct B"; }
    virtual ICloneable* clone() const { return new B; }
};

#include <iostream>
#include <typeinfo>

int main()
{
    A a;
    B b;

    ICloneable* aclone = a.clone();
    ICloneable* bclone = b.clone();

    std::cout << typeid(*aclone).name() << "\n";
    std::cout << typeid(*bclone).name() << "\n";

    delete aclone;
    delete bclone;
}

打印（依赖于编译器）：

1A
1B

Answer 3

如果您选择的语言支持C++ code example返回类型，则可以执行与以下covariant类似的操作：

struct A {
    virtual A *make() = 0;
};

struct B : public A {
    B *make() override {
        return new B{};  
    }
};

虽然这不符合您在基类中定义一次的标准，但我认为值得一提。