如何在派生类中重写模板类的静态方法 [英] How to override static method of template class in derived class

问题描述

我有一些基本的静态替代静态方法，但是整个问题非常复杂且时间太长(游戏引擎中的资源管理概述)，所以这里是一个简化的版本:

I'm having a little issue with overriding static methods of base clases, but the whole question is very complicated and too long (generalization of resource management in game engine), so here's a simplified version:

template<class T>
class base
{
    static void bar()
    { printf("bar"); }
public:
    static void foo()
    { bar(); }
};

class derived : public base<int>
{
    static void bar()
    { printf("baz"); }
};

int main() { derived::foo(); }

在我的情况下，上面的代码输出"bar"，说明我希望它输出"baz".我该怎么办?似乎无论我尝试什么，base :: foo()总是调用base :: bar().我的设计可能有问题.我从来没有遇到过这个问题-我该如何解决?

The code above outputs "bar" in my case, insted I want it to output "baz". How can I go about that? It seems that no matter what I attempt, base::foo() always calls base::bar(). I might have an issue with the design. I've never came across this problem - how can I resolve it?

推荐答案

简单的类继承是无法实现的.一个方法不能同时是static和virtual.

What you're trying to do is not achievable with simple class inheritance; a method cannot be both static and virtual.

您需要一个static方法，以便能够在没有对象(实例)的情况下调用函数；并且您需要bar为virtual，以便bar<int>::foo()从derived 实例调用时会调用derived::bar().

You need a static method to be able to call a function without an object (an instance); and you need bar to be virtual so that bar<int>::foo() calls derived::bar() when called from a derived instance.

这两个特征是互斥的.但是好奇递归模板模式(CRTP)可能是这里的解决方案:

Those two traits are mutually exclusive. But the Curiously Recursive Template Pattern (CRTP) may be a solution here:

#include <iostream>

template<class T>
struct base
{
    static void foo()
    {
        T::bar();
    }
};

struct derived : public base<derived>
{
    static void bar()
    {
        std::cout << "derived" << std::endl;
    }
};

int main()
{
    derived::foo();
}

实时示例

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