java - C++中接口(interface)的匿名实现

Question

C++ 是否支持像 Java 这样的接口(interface)的匿名实现？

类似下面的代码:

Runnable myRunnable =
    new Runnable(){
        public void run(){
            System.out.println("Runnable running");
        }
    }

Answer 1

您可以通过创建一个继承自您定义的接口(interface)的匿名类的实例来非常接近。

例子:

#include <iostream>

struct Runnable {
    virtual ~Runnable() = default;
    virtual void operator()() = 0;
};

int main() {
    struct : Runnable {             // anonymous implementation of interface
        void operator()() override {
            std::cout << "Run Forrest\n";
        }
    } myRunnable;

    myRunnable(); // prints Run Forrest
}

---

---

这是一个人为的例子，将基类指针指向 Runnable 的不同匿名实现的实例在 std::vector<std::unique_ptr<Runnable>> .正如您在此处看到的，使实现匿名化几乎没有用处。匿名类型就是 std::remove_reference_t<decltype(*instance)> .

#include <iostream>
#include <memory>
#include <type_traits>
#include <vector>

// Runnable is defined as in the example above

int main() {
    // a collection of Runnables:
    std::vector<std::unique_ptr<Runnable>> runners;

    { // create a pointer to an instance of one anonymous impl.
        struct : Runnable {
            void operator()() override {
                std::cout << "Foo\n";
            }
        } *instance = new std::remove_reference_t<decltype(*instance)>;
        
        runners.emplace_back(instance); // save for later
    }

    { // create a pointer to an instance of another anonymous impl.
        struct : Runnable {
            void operator()() override {
                std::cout << "Bar\n";
            }
        } *instance = new std::remove_reference_t<decltype(*instance)>;
        
        runners.emplace_back(instance); // save for later
    }    

    // loop through all `Runnable`s and ... run them:
    for(auto& runner : runners) (*runner)(); // prints "Foo\nBar\n"
}