C语言当中的OOP

前言

最近搞 Unix编程 & PHP扩展开发 搞得不亦乐乎，忽然，凭我低下的C语言掌控能力，忽然看到一个不能理解的片段（来自Swoole):

struct _swReactor
{
    /*
         .... 省略100行
        */

    int (*add)(swReactor *, int fd, int fdtype);
    int (*set)(swReactor *, int fd, int fdtype);
    int (*del)(swReactor *, int fd);
    int (*wait)(swReactor *, struct timeval *);
    void (*free)(swReactor *);

    int (*setHandle)(swReactor *, int fdtype, swReactor_handle);
    swDefer_callback *defer_callback_list;

    void (*onTimeout)(swReactor *);
    void (*onFinish)(swReactor *);

    void (*enable_accept)(swReactor *);

    int (*write)(swReactor *, int, void *, int);
    int (*close)(swReactor *, int);
    int (*defer)(swReactor *, swCallback, void *);
};

嗯嗯嗯嗯？ 为何在 struct 里面有函数指针？作者想干什么？ 哦，继续往下看代码，发现，原来是在用C写OOP。

C语言下的面向对象编程

所以，参考了各方教程，写了个小demo:

#include <stdio.h>
#include <stdlib.h>

typedef struct people People;

struct people{
    int age;
    int (*get_age) (People *);
};

int get_age(People *p) {
    return p->age;
}

People base = {
    0,
    get_age
};

People* newPeople(int age)
{
    People *new_people = (People *)malloc(sizeof(People));
    new_people = &base;
    new_people->age = age;
    return new_people;
}

int main() {
    char string[100];
    People *people1 = newPeople(20);
    printf("Get age %d", people1->get_age(people2));
    return 0;
}