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;
}
I have class hierarchy as shown below. It's a simplified version of actual code.
class Base
{
public :
// user_define_type is a output parameter
virtual void Fill(user_define_type);
}
class A : public Base
{
public :
void Fill(user_define_type) override;
}
class B : public Base
{
public :
void Fill(user_define_type) override;
}
I am overriding Fill() method as I need different formatting in both derived classes. Now I have to write one more class deriving from "Base" as it has common functionality. Now my problem is that new class will have to implement Fill() that will operate on different user defined type. As I am returning base class pointer from factory so new Fill() has to be virtual in base but that means I have to add it's definition in older classes "A" and "B" and throw not supported exception from them. This is not a good design. Any better design you guys can suggest ? Thanks in advance.
I believe you need to create a common base class for your user_defined_types in order to achieve this. I also think this could be a good place to use the strategy pattern.
Basically, you create
class user_defined_type_base
{
...
}
class user_defined_type_derived : public user_defined_type_base
{
...
}
class DoSomething
{
private:
DoSomethingStrategy *strategy;
public:
DoSomething(DoSomethingStrategy *strategy) { this->strategy = strategy; }
void Fill(user_defined_type_base *type) { this->strategy->Fill(type); }
}
class DoSomethingStrategy
{
public:
virtual void Fill(user_defined_type_base *obj) = 0;
}
class DoSomethingStrategyA : public DoSomethingStrategy
{
public:
void Fill(user_defined_type_base *obj)
{
...
}
}
class DoSomethingStrategyB : public DoSomethingStrategy
{
public:
void Fill(user_defined_type_base *obj)
{
...
}
}
class DoSomethingStrategyC : public DoSomethingStrategy
{
public:
void Fill(user_defined_type_base *obj)
{
...
}
}
void main()
{
DoSomethingStrategy *strategy = new DoSomethingStragegyA();
DoSomething *dosomething = new DoSomething(strategy);
user_defined_type_base *type = new user_defined_type_base();
dosomething->Fill(type);
DoSomethingStrategy *strategyC = new DoSomethingStragegyC();
DoSomething *dosomethingC = new DoSomething(strategyC);
user_defined_type_base *typeC = new user_defined_type_derived();
dosomethingC->Fill(typeC);
}
class A
{
public:
void doFirstJob()
{
// Do first Job.
}
}
class B : public A
{
public:
virtual void doSecondJob()
{
// Do Second Job.
}
}
class C
{
public:
void doSomething() {
b->doFirstJob();
b->doSecondJob();
}
private:
B* b;
}
Now I should write unit test code for class C, then I'll write a mock for class B, but the problem is how to mock the method doFirstJob().
Bluntly, I want know how to mock the non-virtual method of the parent class???
Can any one help me ??
Typemock Isolator++ supports mocking non virtual methods of a parent class (same as faking a method of the class under test).
See following example:
class A
{
public:
int doFirstJob()
{
return 0;
}
};
class B : public A
{
};
class C
{
public:
int doSomething()
{
return b->doFirstJob();
}
void setB(B* to)
{
b = to;
}
private:
B* b;
};
In the test You create a fake of B -> change the behavior of doFirstJob to return 3 -> continue with your test as you would normally write it.
TEST_CLASS(NonVirtualMethod)
{
public:
TEST_METHOD(NonVirtualMethodTestOfBaseClass)
{
B* fakeB = FAKE<B>();
WHEN_CALLED(fakeB->doFirstJob()).Return(3);
C c;
c.setB(fakeB);
int first = c.doSomething();
Assert::AreEqual(3,first);
}
}
You can find more examples here.
Does NUnit provide functionality to have TestCase attributes associated with a TestFixture? I'd like to run the same set of tests on different implementations of the same interface. The concrete implementation would then be chosen in the SetUp method, and each test should be run with every implementation.
I don't see any possibility for this. But nice idea anyway.
If you are not keen about the TestCase atribute you could use the TestCaseSource attribute with an abstract base class.
public abstract class A
{
public class TestCases : List<TestCaseData>
{
public TestCases()
{
Add(new TestCaseData(1));
Add(new TestCaseData(2));
}
}
public abstract void Test(int i);
}
[TestFixture]
public class B : A
{
[Test]
[TestCaseSource(typeof(TestCases))]
public override void Test(int i)
{
Assert.That(i, Is.GreaterThan(0));
}
}
[TestFixture]
public class C : A
{
[Test]
[TestCaseSource(typeof(TestCases))]
public override void Test(int i)
{
Assert.That(i, Is.LessThan(10));
}
}
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.