C++之函数指针实现函数回调

1、问题

C++很多地方需要函数指针回调，但是我理解还是不够深刻，今天再写了测试例子，方便后面更深入理解和记忆。

2、代码实现

A.h 文件实现

//
//  A.h
//  TestC++
//
//  Created by 1111 on 17/8/17.
//  Copyright © 2017年 sangfor. All rights reserved.
//


#ifndef A_h
#define A_h

#include <iostream>

class A
{
    
public:
    A(void){}
    ~A(void){}
    typedef void(*fun)(int);
    typedef void(*fun1)(void*, int);
    void callback(fun f, int n)
    {
        std::cout << "callback before" << std::endl;
        f(n);
        std::cout << "callback after" << std::endl;
    }
    void callback1(void *obj, fun1 f1, int n1)
    {
        object = obj;
        f = f1;
        n = n1;
    }
    void exec()
    {
        std::cout << "callback1 before" << std::endl;
        f(object, n);
        std::cout << "callback1 before" << std::endl;
    }
private:
    fun1 f;
    void *object;
    int n;
};

#endif /* A_h */

main.cpp文件实现

//
//  main.cpp
//  TestC++
//
//  Created by 1111 on 17/3/19.
//  Copyright © 2017年 sangfor. All rights reserved.
//

#include <iostream>
#include <string.h>
#include "A.h"

using namespace std;

void printf(int n) {
    std::cout << n << std::endl;
}

class B
{
public:
    static void fun(void *p, int n)
    {
        std::cout << "fun" << std::endl;
        B *b = (B *)p;
        b->test(n);
    }
    
    void test(int n)
    {
        if (n % 2 == 0)
            std::cout << n << " % 2 == 0" << std::endl;
        else
            std::cout << n << " % 2 != 0" << std::endl;
    }
    void exec()
    {
        A a;
        a.callback1(this, fun, 100);
        std::cout << "B exec" << std::endl;
        a.exec();
    }
};


int main(int argc, const char * argv[]) {
    printf("hello\n");
    A a;
    a.callback(printf, 5);
    B b;
    b.exec();
    //f(get);
    return 0;
}

3、结果

hello
callback before
5
callback after
B exec
callback1 before
fun
100 % 2 == 0
callback1 before
Program ended with exit code: 0