Question

我想调用类的运行时绑定函数，它继承了公共类“Bindable”的绑定能力。这有可能吗？

这是一个存根肯定缺少很多模板参数和命名空间：

#include <iostream>     // std::cout
#include <functional>   // std::bind
#include <map>          // std::map

class Bindable {
public:
    void bindFunction (int x, auto newFn) {
        mFns.insert(std::pair<int, auto>(x,newFn));
    }
    void invokeFunction (int key) {
        mFns.at(key)();
    }

protected:
    std::map<int, function> mFns;
};

class A : Bindable {
    void funAone (void) {
        cout << "called funAone" <<std::endl;
    }
    void funAtwo (void) {
        cout << "called funAtwo" <<std::endl;
    }
};

class B : Bindable {
    void funBone (void) {
        cout << "called funBone" <<std::endl;
    }
    void funBtwo (void) {
        cout << "called funBtwo" <<std::endl;
    }
};

int main() {
    A a;
    B b;

    a.bindFunction(1, &A::funAone);
    a.bindFunction(2, &A::funAtwo);
    b.bindFunction(1, &B::funBone);
    b.bindFunction(2, &B::funBtwo);

    a.invokeFunction(1);
    a.invokeFunction(2);
    b.invokeFunction(1);
    b.invokeFunction(2);
}

Answer 1

是的，可以使用std :: bind。请注意，auto不能用作函数或模板参数。

#include <iostream>     // std::cout
#include <functional>   // std::bind
#include <map>          // std::map

class Bindable {
public:
    typedef std::function<void()> Function;
    void bindFunction (int x, Function newFn) {
        mFns.insert(std::pair<int, Function>(x,newFn));
    }
    void invokeFunction (int key) {
        mFns.at(key)();
    }

protected:
    std::map<int, Function > mFns;
};

class A : public Bindable {
public:
    void funAone (void) {
        std::cout << "called funAone" <<std::endl;
    }
    void funAtwo (void) {
        std::cout << "called funAtwo" <<std::endl;
    }
};

class B : public Bindable {
public:
    void funBone (void) {
        std::cout << "called funBone" <<std::endl;
    }
    void funBtwo (void) {
        std::cout << "called funBtwo" <<std::endl;
    }
};

int main() {
    A a;
    B b;

    a.bindFunction(1, std::bind(&A::funAone, a)); // more than one way to bind
    a.bindFunction(2, std::bind(&A::funAtwo, &a)); // the object parameter
    b.bindFunction(1, std::bind(&B::funBone, b));
    b.bindFunction(2, std::bind(&B::funBtwo, &b));

    a.invokeFunction(1);
    a.invokeFunction(2);
    b.invokeFunction(1);
    b.invokeFunction(2);
}

Answer 2

选项＃1

使用 CRTP idiom 来了解可以存储成员函数的指针类型：

template <typename T>
struct Bindable {   
    void bindFunction (int x, void(T::*newFn)()) {
        mFns.insert(std::make_pair(x,newFn));
    }
    void invokeFunction (int key) {
        (static_cast<T*>(this)->*mFns.at(key))();
    }

protected:
    std::map<int, void(T::*)()> mFns;
};

struct A : Bindable<A> {
    void funAone (void) {
        std::cout << "called funAone" <<std::endl;
    }
    void funAtwo (void) {
        std::cout << "called funAtwo" <<std::endl;
    }
};

DEMO 1

选项＃2

使用类型擦除并使bindFunction成为函数模板：

struct Bindable {    
    template <typename T, typename std::enable_if<std::is_base_of<Bindable, T>{}, int>::type = 0>
    void bindFunction (int x, void(T::*newFn)()) {
        mFns.insert(std::make_pair(x, std::bind(newFn, static_cast<T*>(this))));
    }
    void invokeFunction (int key) {
        mFns.at(key)();
    }
protected:
    std::map<int, std::function<void()>> mFns;
};

struct A : Bindable {
    void funAone (void) {
        std::cout << "called funAone" <<std::endl;
    }
    void funAtwo (void) {
        std::cout << "called funAtwo" <<std::endl;
    }
};

DEMO 2

在这两种情况下，您都可以使用以下代码：

int main() {
    A a;
    B b;

    a.bindFunction(1, &A::funAone);
    a.bindFunction(2, &A::funAtwo);
    b.bindFunction(1, &B::funBone);
    b.bindFunction(2, &B::funBtwo);

    a.invokeFunction(1);
    a.invokeFunction(2);
    b.invokeFunction(1);
    b.invokeFunction(2);
}

输出：

called funAone
called funAtwo
called funBone
called funBtwo

c ++ 11：std :: bind用于子类成员函数

2 个答案:

选项＃1

选项＃2