a.base1::start(); a.base2::start(); or if you want to use one specifically class derived:public base1,public base2 { public: using base1::start; };
This problem doesn’t come up very often. The solution I’m familiar with was designed by Doug McIlroy and appears in Bjarne Stroustrup’s books (presented in both Design & Evolution of C++ section 12.8 and The C++ Programming Language section 25.6). According to the discussion in Design & Evolution, there was a proposal to handle this … Read more
Multiple inheritance (abbreviated as MI) smells, which means that usually, it was done for bad reasons, and it will blow back in the face of the maintainer. Summary Consider composition of features, instead of inheritance Be wary of the Diamond of Dread Consider inheritance of multiple interfaces instead of objects Sometimes, Multiple Inheritance is the … Read more
In the first example you’re not actually inheriting from two classes, but from one class and an interface. C# doesn’t allow multiple inheritance from classes, but does allow you to implement multiple interfaces. See this MSDN blog post (link is dead so text is pasted below) for more information on why. You will have to … Read more
You have two options here; use interfaces, or use composition. Honestly, interfaces are very powerful, and after reading this line The interface solution doesn’t make any sense to me. (Interfaces have never really made much sense to me, to be honest…) I think that you should learn how to use them properly. That said, if … Read more
Most likely this is a XY problem. But … just don’t call it twice. #include <iostream> class Creature { public: virtual void identify() { std::cout << “I’m a creature” << std::endl; } }; class Swimmer : public virtual Creature { public: virtual void identify() override { Creature::identify(); tell_ability(); std::cout << “I’m a swimmer\n”; } virtual … Read more
Create a third interface I3 extends I1 and I2. Then class A and B both implement I3, and the generic method accepts I3. That’s perhaps the only way to do it.
super is indeed intended for this situation, but it only works if you use it consistently. If the base classes don’t also all use super it won’t work, and unless the method is in object you have to use something like a common base class to terminate the chain of super calls. class FooBase(object): def … Read more
The error message indicates that you have two conflicting metaclasses somewhere in your hierarchy. You need to examine each of your classes and the QT classes to figure out where the conflict is. Here’s some simple example code that sets up the same situation: class MetaA(type): pass class MetaB(type): pass class A: __metaclass__ = MetaA … Read more
TypeScript 2.2 added support for the ECMAscript 2017 mixin pattern Here is a full code sample: // http://justinfagnani.com/2015/12/21/real-mixins-with-javascript-classes/ // https://github.com/Microsoft/TypeScript/wiki/What’s-new-in-TypeScript#support-for-mix-in-classes type Constructor<T> = new (…args: any[]) => T; class S { foo() { console.log(‘foo from S’); } } // Here the possible SuperClass is set to {} (Object) function Mixin1<T extends Constructor<{}>>(SuperClass: T) { return … Read more