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社区首页 >问答首页 >如何理解面向对象的C语言?

如何理解面向对象的C语言?
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Stack Overflow用户
提问于 2018-03-23 06:28:18
回答 2查看 0关注 0票数 0

有没有一组预处理器(ANSI C 89/ISO C 90兼容),它能在C中实现某种(但有用)的对象定向?

EN

Stack Overflow用户

发布于 2018-03-23 15:36:55

您只需使用普通的结构作为对象,并通过指针传递它们:

struct monkey
{
    float age;
    bool is_male;
    int happiness;
};

void monkey_dance(struct monkey *monkey)
{
    /* do a little dance */
}

要获得诸如继承和多态性之类的东西,您必须更加努力地工作。您可以通过让结构的第一个成员成为超类的实例来进行手动继承,然后可以自由地转换到基类和派生类的指针:

struct base
{
    /* base class members */
};

struct derived
{
    struct base super;
    /* derived class members */
};

struct derived d;
struct base *base_ptr = (struct base *)&d;  // upcast
struct derived *derived_ptr = (struct derived *)base_ptr;  // downcast

要获得多态性(即虚拟函数),您可以使用函数指针,也可以选择函数指针表,也称为虚拟表或vtable:

struct base;
struct base_vtable
{
    void (*dance)(struct base *);
    void (*jump)(struct base *, int how_high);
};

struct base
{
    struct base_vtable *vtable;
    /* base members */
};

void base_dance(struct base *b)
{
    b->vtable->dance(b);
}

void base_jump(struct base *b, int how_high)
{
    b->vtable->jump(b, how_high);
}

struct derived1
{
    struct base super;
    /* derived1 members */
};

void derived1_dance(struct derived1 *d)
{
    /* implementation of derived1's dance function */
}

void derived1_jump(struct derived1 *d, int how_high)
{
    /* implementation of derived 1's jump function */
}

/* global vtable for derived1 */
struct base_vtable derived1_vtable =
{
    &derived1_dance, /* you might get a warning here about incompatible pointer types */
    &derived1_jump   /* you can ignore it, or perform a cast to get rid of it */
};

void derived1_init(struct derived1 *d)
{
    d->super.vtable = &derived1_vtable;
    /* init base members d->super.foo */
    /* init derived1 members d->foo */
}

struct derived2
{
    struct base super;
    /* derived2 members */
};

void derived2_dance(struct derived2 *d)
{
    /* implementation of derived2's dance function */
}

void derived2_jump(struct derived2 *d, int how_high)
{
    /* implementation of derived2's jump function */
}

struct base_vtable derived2_vtable =
{
   &derived2_dance,
   &derived2_jump
};

void derived2_init(struct derived2 *d)
{
    d->super.vtable = &derived2_vtable;
    /* init base members d->super.foo */
    /* init derived1 members d->foo */
}

int main(void)
{
    /* OK!  We're done with our declarations, now we can finally do some
       polymorphism in C */
    struct derived1 d1;
    derived1_init(&d1);

    struct derived2 d2;
    derived2_init(&d2);

    struct base *b1_ptr = (struct base *)&d1;
    struct base *b2_ptr = (struct base *)&d2;

    base_dance(b1_ptr);  /* calls derived1_dance */
    base_dance(b2_ptr);  /* calls derived2_dance */

    base_jump(b1_ptr, 42);  /* calls derived1_jump */
    base_jump(b2_ptr, 42);  /* calls derived2_jump */

    return 0;
}
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/-100003675

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