So I want to use a trait class to implement my contract class, is it possible? Because I got some errors when I compile this following code:
code:
class MyContract {
public:
virtual void foo()=0;
};
class TraitClass {
public:
void foo()
{
Serial.println("hello");
}
};
class MyClass : public virtual MyContract, public TraitClass {
// MyClass stuff here.
};
void setup()
{
MyClass* myClass = new MyClass();
myClass->foo();
}
MyClass does not implement the pure virtual method declared in the MyContract superclass. Just because it also inherits from TraitClass doesn't mean that TraitClass's foo() is going to implement it.
You need to connect the dots:
class MyClass : public virtual MyContract, public TraitClass {
void foo() override
{
TraitClass::foo();
}
};
Related
I'm currently trying to wrap my head around the basics of C++ inheritance. Consider the following piece of code:
// Interfaces
class InterfaceBase
{
public:
virtual void SomeMethod() = 0;
};
class InterfaceInherited : public InterfaceBase
{
};
// Classes
class ClassBase : public InterfaceBase
{
public:
virtual void SomeMethod()
{
}
};
class ClassInherited : public ClassBase, public InterfaceInherited
{
};
int main()
{
ClassBase myBase; // OK
ClassInherited myInherited; // Error on this line
return 0;
}
Here I have two interfaces with an inheritance relationship. The same goes for the two classes which implement the interfaces.
This gives me the following compiler error:
C2259 'ClassInherited': cannot instantiate abstract class
It seems that the class ClassInherited does not inherit the implementation of SomeMethod from ClassBase. Thus it is abstract and cannot be instantiated.
How would I need to modify this simple example in order to let ClassInherited inherit all the implemented methods from ClassBase?
You are encountering a diamond problem.
The solution is to use virtual inheritance (Live), to ensure that only one copy of base class members are inherited by grand-childs:
// Interfaces
class InterfaceBase
{
public:
virtual void SomeMethod() = 0;
};
class InterfaceInherited : virtual public InterfaceBase
{
};
// Classes
class ClassBase : virtual public InterfaceBase
{
public:
virtual void SomeMethod()
{
}
};
class ClassInherited : public ClassBase, public InterfaceInherited
{
};
int main()
{
ClassBase myBase; // OK
ClassInherited myInherited; // OK
return 0;
}
Hi I am trying to implement an interface in C++. I want to be able to call a function from a class that could be implemented by various different classes. The approach I have tried fails since I cannot call the function with a pointer to the interface (abstract class). Here is the basic gist of the code I have tried:
Interface class:
class InterfaceClass{
virtual void handle() = 0;
};
Calling class:
CallingClass::CallingClass(InterfaceClass * owner){
this->owner = owner;
}
void CallingClass::doStuff(){
owner->handle();
}
Implementing classes:
class Class1 : public InterfaceClass {
public:
Class1();
void handle();
}
class Class2 : public InterfaceClass {
public:
Class2();
void handle();
}
each of the handle() functions in the implementing classes just prints out the class name. Each implementing class contains an object of the CallingClass which calls doStuff in a separate timer thread. I am trying to keep it so that the CallingClass doesnt need to know anything about the classes that implement the handle() function.
It fails since I cannot call the function of an abstract class. I expected this but cant figure a way around it. Any advise would be much appreciated! Let me know if any more information is needed.
Thanks
you were missing the public: keyword in the Interface class.
you need to remember that C++ class are by default private, so you should add public where it is needed.
this is a working example for you:
class InterfaceClass {
public:
virtual void handle() = 0;
};
class Calling{
public:
Calling(InterfaceClass * owner) {
this->owner = owner;
}
void doStuff() {
owner->handle();
}
~Calling() { delete owner; }
private :
InterfaceClass* owner;
};
class Class1 : public InterfaceClass {
public:
void handle() override
{
std::cout << "Class1"<<std::endl;
}
};
class Class2 : public InterfaceClass {
public:
void handle()
{
std::cout << "Class2" << std::endl;;
}
};
int main(int argc, char** argv)
{
Calling c1(new Class1());
c1.doStuff();
Calling c2(new Class2());
c2.doStuff();
}
I have encountered a virtual method in a nested class.
##classone.h
class ClassOne: {
public:
class InnerClass{
public:
virtual void method1();
...
##classone.cpp
void ClassOne::InnerClass::method1()
{
...
}
I am subclassing ClassOne and need to extend method1(). What need's to be done with the nested class in that situation?
What I tried
##subclassone.h
class SubClassOne: public ClassOne{
public:
virtual void method1();
##subclassone.cpp
void SubClassOne::InnerClass::method1()
{
##New implementation
}
But that gives a multiple definition of ClassOne::InnerClass::method1()
method1 belongs to ClassOne::InnerClass, not ClassOne. When you inherit from ClassOne, the nested class from base class becomes a member of the derived class, too, and you can reach it by qualifying with either ClassOne:: or SubClassOne::. Hence the double definition error regarding method1.
You'll need to sub-class InnerClass, too. If you still wish to derive from ClassOne, it would look like this:
class ClassOne {
public:
class InnerClass {
public:
virtual void method1();
};
};
void ClassOne::InnerClass::method1()
{
}
class SubClassOne : public ClassOne {
class DerivedInnerClass : InnerClass { //
virtual void method1();
};
};
void SubClassOne::DerivedInnerClass::method1()
{
}
I have some trouble.
I made some interface (abstract class in c++). Implemented it in my class (derived from CCObject too).
In third class I try to invoke method of interface and got SIGABORT. Here the code
//interface class
class CallBackInterface
{
public:
virtual void SomeMethod() = 0;
};
//my class that implement the interface
class MyClass : public CallBackInterface, public CCObject
{
public:
void SomeMethod(){/*some realization*/};
};
//class that invoke the SomeMethod
class CallBacker()
{
public:
CallBackInterface* callBackObject;
};
//main code
CallBacker* callBacker = new CallBacker();
MyClass* myClass = new MyClass();
callBacker->callBackObject = myClass;
/*
this string generate unexpected invoke of copyWithZone method CCObject's class
with SIGABORT. */
callBacker->callBackObject->SomeMethod();
/*
In debugger mode I see that SomeMethod don't invoke (debugger don't go into it). Here the copyWithZone*/
CCObject* CCCopying::copyWithZone(CCZone *pZone)
{
CC_UNUSED_PARAM(pZone);
CCAssert(0, "not implement"); <<- here is SIGABORT
return 0;
}
The copyWithZone invokation crashes my app
class CallBackInterface : public CCObject
{
public:
virtual void SomeMethod() = 0;
};
class MyClass : public CallBackInterface
{
void SomeMethod(){}
};
Try this! I met the same problem before.
I have the following class structure:
class InterfaceA
{
virtual void methodA =0;
}
class ClassA : public InterfaceA
{
void methodA();
}
class InterfaceB : public InterfaceA
{
virtual void methodB =0;
}
class ClassAB : public ClassA, public InterfaceB
{
void methodB();
}
Now the following code is not compilable:
int main()
{
InterfaceB* test = new ClassAB();
test->methodA();
}
The compiler says that the method methodA() is virtual and not implemented. I thought that it is implemented in ClassA (which implements the InterfaceA).
Does anyone know where my fault is?
That is because you have two copies of InterfaceA. See this for a bigger explanation: https://isocpp.org/wiki/faq/multiple-inheritance (your situation is similar to 'the dreaded diamond').
You need to add the keyword virtual when you inherit ClassA from InterfaceA. You also need to add virtual when you inherit InterfaceB from InterfaceA.
Virtual inheritance, which Laura suggested, is, of course, the solution of the problem. But it doesn't end up in having only one InterfaceA. It has "side-effects" too, for ex. see https://isocpp.org/wiki/faq/multiple-inheritance#mi-delegate-to-sister. But if get used to it, it may come in handy.
If you don't want side effects, you may use template:
struct InterfaceA
{
virtual void methodA() = 0;
};
template<class IA>
struct ClassA : public IA //IA is expected to extend InterfaceA
{
void methodA() { 5+1;}
};
struct InterfaceB : public InterfaceA
{
virtual void methodB() = 0;
};
struct ClassAB
: public ClassA<InterfaceB>
{
void methodB() {}
};
int main()
{
InterfaceB* test = new ClassAB();
test->methodA();
}
So, we are having exactly one parent class.
But it looks more ugly when there is more than one "shared" class (InterfaceA is "shared", because it is on top of "dreaded diamond", see here https://isocpp.org/wiki/faq/multiple-inheritance as posted by Laura). See example (what will be, if ClassA implements interfaceC too):
struct InterfaceC
{
virtual void methodC() = 0;
};
struct InterfaceD : public InterfaceC
{
virtual void methodD() = 0;
};
template<class IA, class IC>
struct ClassA
: public IA //IA is expected to extend InterfaceA
, public IC //IC is expected to extend InterfaceC
{
void methodA() { 5+1;}
void methodC() { 1+2; }
};
struct InterfaceB : public InterfaceA
{
virtual void methodB() = 0;
};
struct ClassAB
: public ClassA<InterfaceB, InterfaceC> //we had to modify existing ClassAB!
{
void methodB() {}
};
struct ClassBD //new class, which needs ClassA to implement InterfaceD partially
: public ClassA<InterfaceB, InterfaceD>
{
void methodB() {}
void methodD() {}
};
The bad thing, that you needed to modify existing ClassAB. But you can write:
template<class IA, class IC = interfaceC>
struct ClassA
Then ClassAB stays unchanged:
struct ClassAB
: public ClassA<InterfaceB>
And you have default implementation for template parameter IC.
Which way to use is for you to decide. I prefer template, when it is simple to understand. It is quite difficult to get into habit, that B::incrementAndPrint() and C::incrementAndPrint() will print different values (not your example), see this:
class A
{
public:
void incrementAndPrint() { cout<<"A have "<<n<<endl; ++n; }
A() : n(0) {}
private:
int n;
};
class B
: public virtual A
{};
class C
: public virtual A
{};
class D
: public B
: public C
{
public:
void printContents()
{
B::incrementAndPrint();
C::incrementAndPrint();
}
};
int main()
{
D d;
d.printContents();
}
And the output:
A have 0
A have 1
This problem exists because C++ doesn't really have interfaces, only pure virtual classes with multiple inheritance. The compiler doesn't know where to find the implementation of methodA() because it is implemented by a different base class of ClassAB. You can get around this by implementing methodA() in ClassAB() to call the base implementation:
class ClassAB : public ClassA, public InterfaceB
{
void methodA()
{
ClassA::methodA();
}
void methodB();
}
You have a dreaded diamond here.
InterfaceB and ClassA must virtually inherit from InterfaceA
Otherwise you ClassAB has two copies of MethodA one of which is still pure virtual. You should not be able to instantiate this class. And even if you were - compiler would not be able to decide which MethodA to call.