Define a method outside of a class with explicit interface.

Question

0.00/5 (No votes)

See more:

Hello everybody. I am trying to implement two abstract methods (same name, same signatures). The problem is, I can define a method inside the class, but I cant (or I dont know how to) define it outside the class.
Is my syntax bad or it cannot be done this way in C++?
Could you please help me? Thank you!

What I have tried:

This is my sample code:

C++

interface IA
{
public:
	virtual void Test()=0;
};

interface IB
{
public:
	virtual void Test()=0;
};

class CTest:public IA, public IB
{
public:
	void IA::Test();
	void IB::Test();
};

void CTest::IA::Test()			// error C2509
{
}

But this one works like a charm (but I dont want internal definitions):

C++

class CTest:public IA, public IB
{
public:
	void IA::Test() {cout << "Test from A" << endl;}
	void IB::Test() {cout << "Test from B" << endl;}
};

Posted 23-Aug-22 2:48am

Member 13749758

Updated 23-Aug-22 3:27am

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

CPallini · Answer 1 · 2022-08-23T03:27:00

Solution 1

What's wrong with

C++

class IA
{
public:
  virtual void Test()=0;
};

class IB
{
public:
  virtual void Test()=0;
};

class CTest:public IA, public IB
{
public:
  void Test() override;
};

void CTest::Test()
{
}

?

Posted 23-Aug-22 3:27am

CPallini

Comments

Rick York 23-Aug-22 9:52am

Doesn't this have a diamond problem? https://www.cprogramming.com/tutorial/virtual_inheritance.html

CPallini 23-Aug-22 9:58am

Nope. In the above code, the compiler exactly knows which is the method to call. Try:


using namespace std;
class IA
{
public:
  virtual void Test()=0;
};

class IB
{
public:
  virtual void Test()=0;
};

class CTest:public IA, public IB
{
public:
  void Test() override;
};

void CTest::Test()
{
  cout << "Test\n";
}

int main()
{
  CTest ct;

  IA * pA = &ct;
  IB * pB = &ct;

  ct.Test();
  pA->Test();
  pB->Test();

}

Greg Utas 23-Aug-22 10:47am

Test overrides IA::Test and IB::Test simultaneously. Defining functions with the same name in base classes that will be multiply inherited seems dubious to me, but maybe there are legitimate examples.

Member 13749758 23-Aug-22 15:50pm

Thanks for your answer. I've already tried similar way. But the problem is that in my first sample, I have two methods with same name, same signatures, distinguished by scope resolution operator, but I cant define these methods outside class CTest.
I just wonder if it is possible. I dont want one method to override two virtual methods (as in your reply). I want one method override the one from IA and second method override the one from IB.

This one works:
class CTest:public IA, public IB
{
public:
void IA::Test() {cout << "Test from A" << endl;}
void IB::Test() {cout << "Test from B" << endl;}
};

But I want it to be outside the class...