是否可以使用CRTP在静态多态中模拟纯虚函数? [英] Is emulating pure virtual function in static polymorphism using CRTP possible?

问题描述

我正在尝试使用CRTP实现编译时多态性，并希望强制派生类实现该功能.

I'm trying to implement compile-time polymorphism using CRTP, and want to force the derived class to implement the function.

当前的实现是这样的.

template <class Derived>
struct base {
    void f() {
        static_cast<Derived*>(this)->f();
    }
};

struct derived : base<derived>
{
    void f() {
    ...
    }
};

在此实现中，如果派生类未实现 f()，则对该函数的调用会陷入无限循环.

In this implementation, call to the function falls into an infinite loop if the derived class didn't implement f().

如何强制派生类实现类似于纯虚函数的函数?我尝试使用像 static_assert(& base :: f！=& Derived :: f，"...")这样的"static_assert"，但是它会生成一条错误消息，指出两个成员函数指针指向不同类的成员函数是不可比的.

How do I force the derived class to implement the function like pure virtual function? I tried to use 'static_assert' like static_assert(&base::f != &Derived::f, "...") but it generates an error message saying that two member function pointers pointing to the different classes' member functions are not comparable.

推荐答案

您可以为您覆盖的内容和钩子赋予不同的名称，如下所示:

You can give the thing you override and the hook different names, like this:

template <class Derived>
struct base {
    void f() {
        static_cast<Derived*>(this)->fimpl();
    }
    void fimpl() = delete;
};

struct derived : base<derived> {
    void fimpl() { printf("hello world\n"); }
};

在这里， fimpl = delete 在基部中，因此除非在派生类中覆盖了 fimpl ，否则它不会被意外调用.

Here, fimpl = delete in the base so that it cannot be called accidentally unless fimpl is overridden in the derived class.

您还可以将中间隐藏层粘贴到CRTP中，以临时"将 f 标记为 delete :

You can also stick an intermediate hiding layer into your CRTP to "temporarily" mark f as delete:

template <class Derived>
struct base {
    void f() {
        static_cast<Derived*>(this)->f();
    }
};

template <class Derived>
struct intermediate : base<Derived> {
    void f() = delete;
};

struct derived : intermediate<derived> {
    void f() { printf("hello world\n"); }
};

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