C++11 std::bind std::function 高级使用方法

从最基础的了解，std::bind和std::function

/*
 * File:   main.cpp
 * Author: Vicky.H
 * Email:  eclipser@163.com
 */
#include <iostream>
#include <functional>
#include <typeinfo>
#include <string.h>

int add1(int i, int j, int k) {
    return i + j + k;
}

class Utils {
public:
    Utils(const char* name) {
        strcpy(_name, name);
    }

    void sayHello(const char* name) const {
        std::cout << _name << " say: hello " << name << std::endl;
    }

    static int getId() {
        return 10001;
    } 

    int operator()(int i, int j, int k) const {
        return i + j + k;
    }

private:
    char _name[32];
};

/*
 *
 */
int main(void) {

    // 绑定全局函数
    auto add2 = std::bind(add1, std::placeholders::_1, std::placeholders::_2, 10);
    // 函数add2 = 绑定add1函数。參数1不变，參数2不变。參数3固定为10.
    std::cout << typeid(add2).name() << std::endl;
    std::cout << "add2(1,2) = " << add2(1, 2) << std::endl;

    std::cout << "\n---------------------------" << std::endl;

    // 绑定成员函数
    Utils utils("Vicky");
    auto sayHello = std::bind(&Utils::sayHello, utils/*调用者*/, std::placeholders::_1/*參数1*/);
    sayHello("Jack");

    auto sayHelloToLucy = std::bind(&Utils::sayHello, utils/*调用者*/, "Lucy"/*固定參数1*/);
    sayHelloToLucy();

    // 绑定静态成员函数
    auto getId = std::bind(&Utils::getId);
    std::cout << getId() << std::endl;

    std::cout << "\n---------------------------" << std::endl;

    // 绑定operator函数
    auto add100 = std::bind(&Utils::operator (), utils, std::placeholders::_1, std::placeholders::_2, 100);
    std::cout << "add100(1, 2) = " << add100(1, 2) << std::endl;

    // 注意：无法使用std::bind()绑定一个重载函数

    return 0;
}

/*
 * File:   main2.cpp
 * Author: Vicky.H
 * Email:  eclipser@163.com
 */
#include <iostream>
#include <typeinfo>

void sayHello() {
    std::cout << "Hello world !" << std::endl;
}

int sum(int i, int j, int k) {
    return i + j + k;
}

template <typename T>
class Func {
public:

    Func(T fun) {
        if (!fun) {
            throw "fun nullptr";
        }
        _fun = fun;
    }

    template<typename R, typename A1, typename A2, typename A3, typename A4, typename A5>
    R Call(A1 a1, A2 a2, A3 a3, A4 a4, A5 a5) {
        return _fun(a1, a2, a3, a4, a5);
    }

    template<typename R, typename A1, typename A2, typename A3, typename A4>
    R Call(A1 a1, A2 a2, A3 a3, A4 a4) {
        return _fun(a1, a2, a3, a4);
    }

    template<typename R, typename A1, typename A2, typename A3>
    R Call(A1 a1, A2 a2, A3 a3) {
        return _fun(a1, a2, a3);
    }

    template<typename R, typename A1, typename A2>
    R Call(A1 a1, A2 a2) {
        return _fun(a1, a2);
    }

    template<typename R, typename A1>
    R Call(A1 a1) {
        return _fun(a1);
    }

    template<typename R>
    R Call() {
        return _fun();
    }

    void Call() {
        _fun();
    }

private:
    T _fun;
};

#include <functional>

template<typename R = void, typename... Args>
class Fn {
public:
    Fn(std::function<R(Args...)> fun) : _fun(fun) {
    }

    R operator()(Args... args) {
        return _fun(args...);
    }
private:
    std::function<R(Args...) > _fun;
};

/*
 * 将函数注冊到对象中。通过对象直接调用
 */
int main(void) {

    Func<void(*)() > sayHelloFunc(sayHello);
    sayHelloFunc.Call();

    Func<int (*)(int, int, int) > sumFunc(sum);
    std::cout << "sumFunc.Call<int>(1, 2, 3) : " << sumFunc.Call<int>(1, 2, 3) << std::endl;

    std::cout << "\n---------------------------" << std::endl;

    Fn<> sayHelloFn(sayHello);
    sayHelloFn();

    Fn<int, int, int, int> sumFn(sum);
    std::cout << "sumFn(1, 2, 3) : " << sumFn(1, 2, 3) << std::endl;

    std::cout << "\n---------------------------" << std::endl;

    return 0;
}

Hello world !

sumFunc.Call<int>(1, 2, 3) : 6





---------------------------

Hello world !

sumFn(1, 2, 3) : 6





---------------------------

上面的样例很有趣，使用了2种方案。将一个函数，注冊到一个对象/仿函数中，而且通过一个对象/仿函数来直接调用调用。

样例显而易见的。第2种方案更佳简洁，而且对传递參数有明白的推断，当參数类型或数量不对的时候，编译器将导致失败。

这样的方案，能够将类的成员变量直接作为函数的參数使用，或者，如我：

http://blog.csdn.net/eclipser1987/article/details/23926395

这篇文章中，无法直接调用脚本函数类。有了好的解决的方法。这个我将随后补充。

#include <list>
#include <functional>

template<typename... Args>
class Fns
{
private:

	std::list<std::function<void(Args...)> > _calls;

public:

	virtual ~Fns()
	{
		_calls.clear();
	}

	void connect(std::function<void(Args...)> fct)
	{
		_calls.push_back(fct);
	}

	template<typename Object>
	void connect(Object* object, void (Object::*method)(Args...))
	{
		_calls.push_back([object,method](Args... args){(*object.*method)(args...);});
	}

	template<typename Object>
	void connect(Object* object, void (Object::*method)(Args...) const)
	{
		_calls.push_back([object,method](Args... args){(*object.*method)(args...);});
	}

	template<typename Object>
	void connect(const Object* object, void (Object::*method)(Args...) const)
	{
		_calls.push_back([object,method](Args... args){(*object.*method)(args...);});
	}

	void emit(Args... args)
	{
		for(auto call : _calls)
			call(args...);
	}
};

#include <cstdio>
#include "Signal.hpp"

class Foo
{
public:

	void bar(int x, int y)
	{
		printf("Foo::bar(%d, %d)\n", x, y);
	}
};

void foobar(int x, int y)
{
	printf("foobar(%d, %d)\n", x, y);
}

int main(void)
{
	Foo foo;
	Fns<int, int> s;

	// Connect a function
	s.connect(foobar);
	// Connect a class method
	s.connect(&foo, &Foo::bar);
	// Create and connect some lambda expression
	s.connect([&foo](int x, int y){
		printf("lambda::"); foo.bar(x, y);
	});
	// Emit the signal !
	s.emit(4, 2);
	getchar();
	return 0;
}

foobar(4, 2)

Foo::bar(4, 2)

lambda::Foo::bar(4, 2)