c++ - 基类方法的接口(interface)

Question

对于下面的代码，还有其他方法可以通过接口(interface)访问基类中的方法吗？

struct Base {
    void funct_base() {
        printf("Common function for class Foo and class Bar\n");
    }
};

struct IFoo {
    virtual ~IFoo() {}
    virtual void funct_a() = 0;
    // would like to access Base::bunct_base() from here
};

struct Foo : public Base, public IFoo {
    virtual void funct_a() {
        printf("I am Foo:: funct A\n");
    }
};

class IBar {
    virtual ~IBar() {}
    virtual void funct_a() = 0;
    // would like to access Base::bunct_base() from here
};

class Bar : public Base, public IBar {
    virtual void funct_a() {
        printf("I am Bar:: funct A\n");
    }
};

我知道这可以做到，但我就是不喜欢 wrapper ，它看起来不干净:

struct IBar {
    virtual ~IBar() {}
    virtual void funct_a() = 0;
    virtual void funct_base() = 0;
};

struct Bar : public Base {
    virtual void funct_a() {
        printf("I am Bar:: funct A\n");
    }

    virtual void funct_base() {
        Base::funct_base(); 
    }
};

编辑:

问题是，有一个基类，以及继承自同一个基类的两个不同的派生类。有没有办法在不添加基类方法包装器的情况下通过派生类接口(interface)访问基类方法？

Answer 1

使用带有 Abstract method 的抽象基类 IBase funct_base 并将接口(interface)类设为 Virtual base classes Base、IFoo 和 IBar 类:

struct IBase {
    virtual void funct_base() = 0;
};

struct Base : public virtual IBase {
    virtual void funct_base() override {    printf("Common function for class Foo and class Bar\n"); }
};

struct IFoo : public virtual IBase {
    virtual void funct_a() = 0;
};

struct Foo : public IFoo, public Base {
    virtual void funct_a() override { printf("I am Foo:: funct A\n"); }
};

class IBar : public virtual IBase {
    virtual void funct_a() = 0;
};

class Bar : public IBar, public Base {
    virtual void funct_a() override { printf("I am Bar:: funct A\n"); }
};