5

GObject 类 A 实现接口 IA，B 是 A 的派生类。 B 如何覆盖作为接口 IA 一部分的 A 的方法？

或者，这在 GObject 中是否可行？

我知道如何重写父类方法，但是当继承遇到接口时，事情似乎更复杂了。

非常感谢！

4

2 回答 2

5

是的，有可能：只需像第一次一样重新实现接口，在您的函数中使用G_IMPLEMENT_INTERFACE()或手动初始化它。get_type()

真正的痛苦是如果你需要链接旧方法。在这种情况下，你应该玩弄 g_type_interface_peek_parent得到以前的接口类。

这是一个测试用例：

/* gcc -otest `pkg-config --cflags --libs gobject-2.0` test.c */
#include <glib-object.h>


/* Interface */

#define TYPE_IFACE   (iface_get_type())

typedef void Iface;
typedef struct {
    GTypeInterface parent_class;
    void (*action) (Iface *instance);
} IfaceClass;

GType
iface_get_type(void)
{
    static GType type = 0;

    if (G_UNLIKELY(type == 0)) {
        const GTypeInfo info = {
            sizeof(IfaceClass), 0,
        };

        type = g_type_register_static(G_TYPE_INTERFACE, "Iface", &info, 0);
    }

    return type;
}

void
iface_action(Iface *instance)
{
    G_TYPE_INSTANCE_GET_INTERFACE(instance, TYPE_IFACE, IfaceClass)->
        action(instance);
}


/* Base object */

#define TYPE_BASE    (base_get_type())

typedef GObject        Base;
typedef GObjectClass   BaseClass;

static void
base_action(Iface *instance)
{
    g_print("Running base action on a `%s' instance...\n",
            g_type_name(G_TYPE_FROM_INSTANCE(instance)));
}

static void
base_iface_init(IfaceClass *iface)
{
    iface->action = base_action;
}

G_DEFINE_TYPE_WITH_CODE(Base, base, G_TYPE_OBJECT,
                        G_IMPLEMENT_INTERFACE(TYPE_IFACE, base_iface_init));

static void
base_class_init(BaseClass *klass)
{
}

static void
base_init(Base *instance)
{
}


/* Derived object */

#define TYPE_DERIVED (derived_get_type())

typedef Base      Derived;
typedef BaseClass DerivedClass;

static void
derived_action(Iface *instance)
{
    IfaceClass *iface_class, *old_iface_class;

    iface_class = G_TYPE_INSTANCE_GET_INTERFACE(instance, TYPE_IFACE, IfaceClass);
    old_iface_class = g_type_interface_peek_parent(iface_class);

    g_print("Running derived action on a `%s' instance...\n",
            g_type_name(G_TYPE_FROM_INSTANCE(instance)));

    /* Chain up the old method */
    old_iface_class->action(instance);
}

static void
derived_iface_init(IfaceClass *iface)
{
    iface->action = derived_action;
}

G_DEFINE_TYPE_WITH_CODE(Derived, derived, TYPE_BASE,
                        G_IMPLEMENT_INTERFACE(TYPE_IFACE, derived_iface_init));

static void
derived_class_init(DerivedClass *klass)
{
}

static void
derived_init(Derived *instance)
{
}


int
main()
{
    GObject *object;

    g_type_init();

    object = g_object_new(TYPE_BASE, NULL);
    iface_action((Iface *) object);
    g_object_unref(object);

    object = g_object_new(TYPE_DERIVED, NULL);
    iface_action((Iface *) object);
    g_object_unref(object);

    return 0;
}
于 2009-10-22T15:04:52.857 回答
5

我认为更好的解决方案是使 A 的方法虚拟化，而不是让 B 重新实现 A 所附加的接口（这可能需要更多的工作，而不仅仅是重新定义一个函数），你可以这样做（示例应该是完整的除了 fooable 接口定义）：

#include <glib-object.h>
#include "fooable.h"

typedef struct {GObject parent;} A;
typedef struct {
    GObjectClass parent;
    gint (*foo) (Fooable *self, gdouble quux);
} AClass;

#define TYPE_A           (a_get_type())
#define A_CLASS(cls)     (G_TYPE_CHECK_CLASS_CAST((cls), TYPE_A, AClass))
#define A_GET_CLASS(obj) (G_TYPE_INSTANCE_GET_CLASS((obj), TYPE_A, AClass))

gint a_foo_real (Fooable *self, gdouble quux) {
    g_print("a_foo_real(%g)\n", quux);
    return 5;
}

gint a_foo (Fooable *self, gdouble quux) {
    return A_GET_CLASS(self)->foo(self, quux);
}

void implement_fooable (FooableIface *iface) {iface->foo = a_foo;}
void a_class_init      (AClass *cls)         {cls->foo = a_foo_real;}
void a_init            (A *self)             {}

G_DEFINE_TYPE_WITH_CODE(A, a, G_TYPE_OBJECT,
    G_IMPLEMENT_INTERFACE(TYPE_FOOABLE, implement_fooable));

/* derive class B from A  */
typedef struct {A parent;} B;
typedef struct {AClass parent;} BClass;

#define TYPE_B (b_get_type())

gint b_foo_real (Fooable *self, gdouble quux) {
    g_print("b_foo_real(%g)\n", quux);
    return 55;
}

void b_class_init (BClass *cls) {A_CLASS(cls)->foo = b_foo_real;}
void b_init       (B *self)     {}

G_DEFINE_TYPE(B, b, TYPE_A);

int main () {
    g_type_init();
    A *a = g_object_new(TYPE_A, NULL);
    B *b = g_object_new(TYPE_B, NULL);
    fooable_foo(FOOABLE(a), 87.0); // a_foo_real(87.0) and returns 5
    fooable_foo(FOOABLE(b), 32.0); // b_foo_real(32.0) and returns 55
    return 0;
}

这是我能做到的一个简短的例子。当您调用fooable_foo()该函数时，将为您实现接口时定义的函数查看其 vtable，该接口a_foo()查看 A 类的 vtable 以确定实际调用哪个函数。B 类定义a_foo_real()用它自己的覆盖 A 类的定义。如果您需要 B 类的b_foo_real链接，这很容易（使用A_CLASS(b_parent_class)->foo()在 G_DEFINE_TYPE 宏中为您定义）

于 2010-02-16T17:19:16.970 回答